Post on 09-Jul-2020
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Chapter 8 Orthogonality8.3 Diagonalizing Symmetric Matrices and QR
Factorization
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Remembering Back ...
What did it mean for vectors to be orthogonal?
How did we test to see if vectors are orthogonal?
What did it mean for vectors to be orthonormal?
Anyone want to take a shot at defining anorthogonal matrix? It is not a trick question, but not asstraight forward as we may want ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Remembering Back ...
What did it mean for vectors to be orthogonal?
How did we test to see if vectors are orthogonal?
What did it mean for vectors to be orthonormal?
Anyone want to take a shot at defining anorthogonal matrix? It is not a trick question, but not asstraight forward as we may want ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Remembering Back ...
What did it mean for vectors to be orthogonal?
How did we test to see if vectors are orthogonal?
What did it mean for vectors to be orthonormal?
Anyone want to take a shot at defining anorthogonal matrix? It is not a trick question, but not asstraight forward as we may want ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Remembering Back ...
What did it mean for vectors to be orthogonal?
How did we test to see if vectors are orthogonal?
What did it mean for vectors to be orthonormal?
Anyone want to take a shot at defining anorthogonal matrix? It is not a trick question, but not asstraight forward as we may want ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0?
Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal.
Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix?
Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is
7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
DefinitionA square matrix whose columns form an orthonormal basis iscalled an orthogonal matrix.
Example
Is A =
[47
√33√
337 −4
]orthogonal?
Is the dot product of the columns equal to 0? Even so, thesecond vector is not a unit vector, so the matrix is notorthogonal. Is there anything we could do to this matrix toproduce an orthogonal matrix? Divide both entries of thesecond column by the norm, which is 7.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
[12
√32
2√5− 1√
5
]orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors 0?
The columns do not have a dot product of 0, so we do nothave an orthogonal matrix.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
[12
√32
2√5− 1√
5
]orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors 0?
The columns do not have a dot product of 0, so we do nothave an orthogonal matrix.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
[12
√32
2√5− 1√
5
]orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors 0?
The columns do not have a dot product of 0, so we do nothave an orthogonal matrix.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
[12
√32
2√5− 1√
5
]orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors 0?
The columns do not have a dot product of 0, so we do nothave an orthogonal matrix.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
3√11
√211
−√
211
3√11
orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors equal to 0?
Since both are satisfied, the matrix is indeed orthogonal.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
3√11
√211
−√
211
3√11
orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors equal to 0?
Since both are satisfied, the matrix is indeed orthogonal.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
3√11
√211
−√
211
3√11
orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors equal to 0?
Since both are satisfied, the matrix is indeed orthogonal.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
Example
Is A =
3√11
√211
−√
211
3√11
orthogonal?
Are the columns unit vectors?
Is the dot product of the vectors equal to 0?
Since both are satisfied, the matrix is indeed orthogonal.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Why We Care
We looked in chapter 6 at diagonalizing a matrix - when isthat possible?
Here, we want to see if we can orthogonally diagonalize amatrix. That is, see if we can find an orthogonal matrix Pand a diagonal matrix D such that A = PDP−1.
This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part? Finding the inverse ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Why We Care
We looked in chapter 6 at diagonalizing a matrix - when isthat possible?
Here, we want to see if we can orthogonally diagonalize amatrix. That is, see if we can find an orthogonal matrix Pand a diagonal matrix D such that A = PDP−1.
This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part? Finding the inverse ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Why We Care
We looked in chapter 6 at diagonalizing a matrix - when isthat possible?
Here, we want to see if we can orthogonally diagonalize amatrix. That is, see if we can find an orthogonal matrix Pand a diagonal matrix D such that A = PDP−1.
This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part?
Finding the inverse ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Why We Care
We looked in chapter 6 at diagonalizing a matrix - when isthat possible?
Here, we want to see if we can orthogonally diagonalize amatrix. That is, see if we can find an orthogonal matrix Pand a diagonal matrix D such that A = PDP−1.
This is desirable because of what it does for us if we have anorthogonal matrix. Thinking back, though, to finding thediagonalization of a matrix larger than 2× 2, what was thehardest part? Finding the inverse ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices?
It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result?
1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?
Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P?
0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why?
Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to?
The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonal Matrices
TheoremIf P is an orthogonal matrix, then P−1 = PT .
Proof.Let P be an orthogonal matrix and consider PTP.
I How would we find this product of matrices? It wouldbe the dot products of the columns of P.
I If we take the dot product of the i th column of P withthe i th column of P, what will be the result? 1 - why?Because of normality ...
I What will be the product of the i th column of P withthe j th column of P? 0 - why? Because the columns areorthogonal.
I So what is PTP equal to? The identity matrix.
So, PTP = In = P−1P ⇒ PT = P−1.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing by
finding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣
= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[−2 22 1
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ −2− λ 22 1− λ
∣∣∣∣= (−2− λ)(1− λ)− 4 = 0
λ2 + λ− 6 = 0
λ = −3, 2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Now, we find the eigenvectors:
For λ = 2, we have
A− 2I =
[−4 22 −1
]∼[
1 −12
0 0
]
So the eigenvector associated with λ = 2 is
[12
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Now, we find the eigenvectors:For λ = 2, we have
A− 2I =
[−4 22 −1
]∼[
1 −12
0 0
]
So the eigenvector associated with λ = 2 is
[12
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Now, we find the eigenvectors:For λ = 2, we have
A− 2I =
[−4 22 −1
]
∼[
1 −12
0 0
]
So the eigenvector associated with λ = 2 is
[12
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Now, we find the eigenvectors:For λ = 2, we have
A− 2I =
[−4 22 −1
]∼[
1 −12
0 0
]
So the eigenvector associated with λ = 2 is
[12
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
Now, we find the eigenvectors:For λ = 2, we have
A− 2I =
[−4 22 −1
]∼[
1 −12
0 0
]
So the eigenvector associated with λ = 2 is
[12
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
For λ = −3, we have
A + 3I =
[1 22 4
]∼[
1 20 0
]
So the eigenvector associated with λ = −3 is
[−21
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
For λ = −3, we have
A + 3I =
[1 22 4
]
∼[
1 20 0
]
So the eigenvector associated with λ = −3 is
[−21
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
For λ = −3, we have
A + 3I =
[1 22 4
]∼[
1 20 0
]
So the eigenvector associated with λ = −3 is
[−21
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
For λ = −3, we have
A + 3I =
[1 22 4
]∼[
1 20 0
]
So the eigenvector associated with λ = −3 is
[−21
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
So, we have {[12
],
[−21
]}What do we have?
The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.
And how do we normalize?
p1 =v1‖v1‖
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
So, we have {[12
],
[−21
]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because
the vectorsneed to be normalized.
And how do we normalize?
p1 =v1‖v1‖
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
So, we have {[12
],
[−21
]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.
And how do we normalize?
p1 =v1‖v1‖
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
So, we have {[12
],
[−21
]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.
And how do we normalize?
p1 =v1‖v1‖
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
So, we have {[12
],
[−21
]}What do we have? The vectors form an orthogonal basis, sosince we have distinct eigenvalues, we could use this matrixto diagonalize A, but we aren’t done yet because the vectorsneed to be normalized.
And how do we normalize?
p1 =v1‖v1‖
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
p1 =
1√5
[12
]p2 =
1√5
[−21
]So, we have
PTAP =1√5
[1 2−2 1
] [−2 22 1
]1√5
[1 −22 1
]=
[2 00 −3
]= D
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
p1 =1√5
[12
]
p2 =1√5
[−21
]So, we have
PTAP =1√5
[1 2−2 1
] [−2 22 1
]1√5
[1 −22 1
]=
[2 00 −3
]= D
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
p1 =1√5
[12
]p2 =
1√5
[−21
]So, we have
PTAP =1√5
[1 2−2 1
] [−2 22 1
]1√5
[1 −22 1
]=
[2 00 −3
]= D
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
p1 =1√5
[12
]p2 =
1√5
[−21
]
So, we have
PTAP =1√5
[1 2−2 1
] [−2 22 1
]1√5
[1 −22 1
]=
[2 00 −3
]= D
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Orthogonally Diagonalizable Matrices
p1 =1√5
[12
]p2 =
1√5
[−21
]So, we have
PTAP =1√5
[1 2−2 1
] [−2 22 1
]1√5
[1 −22 1
]=
[2 00 −3
]= D
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing by
finding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣
= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Example
Find an orthogonal matrix P that orthogonally diagonalizes
A =
[13 66 −3
]
We begin as we did before when we were diagonalizing byfinding the eigenvalues and associated eigenvectors.
|A− λI | =
∣∣∣∣ 13− λ 66 −3− λ
∣∣∣∣= (13− λ)(−3− λ)− 36 = 0
λ2 − 10λ− 75 = 0
λ = −5, 15
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]
∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]
and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
Now, the normalized vectors:
A− 15I =
[−2 66 −18
]∼[
1 −30 0
]
So, the eigenvector associated with λ = 15 is
[31
]and the
normalized vector is
[3√101√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
For λ = −5, we have
A + 5I =
[18 66 2
]
∼[
1 13
0 0
]
So, the eigenvector associated with λ = −5 is
[1−3
]and the
normalized vector is
[1√10
− 3√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
For λ = −5, we have
A + 5I =
[18 66 2
]∼[
1 13
0 0
]
So, the eigenvector associated with λ = −5 is
[1−3
]and the
normalized vector is
[1√10
− 3√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
For λ = −5, we have
A + 5I =
[18 66 2
]∼[
1 13
0 0
]
So, the eigenvector associated with λ = −5 is
[1−3
]
and the
normalized vector is
[1√10
− 3√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
For λ = −5, we have
A + 5I =
[18 66 2
]∼[
1 13
0 0
]
So, the eigenvector associated with λ = −5 is
[1−3
]and the
normalized vector is
[1√10
− 3√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
For λ = −5, we have
A + 5I =
[18 66 2
]∼[
1 13
0 0
]
So, the eigenvector associated with λ = −5 is
[1−3
]and the
normalized vector is
[1√10
− 3√10
].
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
So, we have
D =
[−5 00 15
]P =
[1√10
3√10
− 3√10
1√10
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
So, we have
D =
[−5 00 15
]P =
[1√10
3√10
− 3√10
1√10
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Another Example
So, we have
D =
[−5 00 15
]P =
[1√10
3√10
− 3√10
1√10
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Example
Find the orthogonal diagonalization of
A =
11 −1 −1−1 11 −1−1 −1 11
|A− λI | =
∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ
∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0
λ = 9, 12, 12
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Example
Find the orthogonal diagonalization of
A =
11 −1 −1−1 11 −1−1 −1 11
|A− λI | =
∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ
∣∣∣∣∣∣
= 1296− 360λ+ 33λ2 − λ3 = 0
λ = 9, 12, 12
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Example
Find the orthogonal diagonalization of
A =
11 −1 −1−1 11 −1−1 −1 11
|A− λI | =
∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ
∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0
λ = 9, 12, 12
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Example
Find the orthogonal diagonalization of
A =
11 −1 −1−1 11 −1−1 −1 11
|A− λI | =
∣∣∣∣∣∣11− λ −1 −1−1 11− λ −1−1 −1 11− λ
∣∣∣∣∣∣= 1296− 360λ+ 33λ2 − λ3 = 0
λ = 9, 12, 12
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Now the normalized vectors: For λ = 9
A− 9I =
2 −1 −1−1 2 −1−1 1 2
∼
1 0 −10 1 −10 0 0
So, the eigenvector associated with λ = 9 is
111
and so the
normalized vector is
1√31√31√3
.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Now the normalized vectors: For λ = 9
A− 9I =
2 −1 −1−1 2 −1−1 1 2
∼1 0 −1
0 1 −10 0 0
So, the eigenvector associated with λ = 9 is
111
and so the
normalized vector is
1√31√31√3
.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Now the normalized vectors: For λ = 9
A− 9I =
2 −1 −1−1 2 −1−1 1 2
∼1 0 −1
0 1 −10 0 0
So, the eigenvector associated with λ = 9 is
111
and so the
normalized vector is
1√31√31√3
.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
For λ = 12, what are we expecting?
2 vectors in the bases ...
A− 12I =
−1 −1 −1−1 −1 −1−1 −1 −1
∼1 1 1
0 0 00 0 0
The eigenvectors associated with λ = 12 are−1
10
,−1
01
and so the normalized vectors are−
1√2
1√2
0
,−
1√2
01√2
.
Are we good?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
For λ = 12, what are we expecting? 2 vectors in the bases ...
A− 12I =
−1 −1 −1−1 −1 −1−1 −1 −1
∼
1 1 10 0 00 0 0
The eigenvectors associated with λ = 12 are−1
10
,−1
01
and so the normalized vectors are−
1√2
1√2
0
,−
1√2
01√2
.
Are we good?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
For λ = 12, what are we expecting? 2 vectors in the bases ...
A− 12I =
−1 −1 −1−1 −1 −1−1 −1 −1
∼1 1 1
0 0 00 0 0
The eigenvectors associated with λ = 12 are−1
10
,−1
01
and so the normalized vectors are−
1√2
1√2
0
,−
1√2
01√2
.
Are we good?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
For λ = 12, what are we expecting? 2 vectors in the bases ...
A− 12I =
−1 −1 −1−1 −1 −1−1 −1 −1
∼1 1 1
0 0 00 0 0
The eigenvectors associated with λ = 12 are−1
10
,−1
01
and so the normalized vectors are−
1√2
1√2
0
,−
1√2
01√2
.
Are we good?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
For λ = 12, what are we expecting? 2 vectors in the bases ...
A− 12I =
−1 −1 −1−1 −1 −1−1 −1 −1
∼1 1 1
0 0 00 0 0
The eigenvectors associated with λ = 12 are−1
10
,−1
01
and so the normalized vectors are−
1√2
1√2
0
,−
1√2
01√2
.
Are we good?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
−
1√2
1√2
0
,−
1√2
01√2
We have a basis, but these vectors are not orthogonal. Sohow do we handle this?
Gram-Schmidt ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
−
1√2
1√2
0
,−
1√2
01√2
We have a basis, but these vectors are not orthogonal. Sohow do we handle this?
Gram-Schmidt ...
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
v1 =
−1√2
1√2
0
v2 =
−1√2
01√2
− projv1
−
1√2
01√2
=
−1√2
01√2
− 12
1
−1√2
1√2
0
=
−1
2√2
− 12√2
1√2
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Now, we need to normalize this vector again.
1√34
−1
2√2
− 12√2
1√2
=2√3
−1
2√2
− 12√2
1√2
=
−1√6
− 1√6
2√6
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
Now, we need to normalize this vector again.
1√34
−1
2√2
− 12√2
1√2
=2√3
−1
2√2
− 12√2
1√2
=
−1√6
− 1√6
2√6
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
One More of These
So, we have
P =
1√3− 1√
2− 1√
61√3
1√2− 1√
61√3
0 2√6
D =
9 0 00 12 00 0 12
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
What Do You Notice ...
If we look at the matrices we used for the last 3 examples,what do they have in common?
[−2 22 1
] [13 66 −3
] 11 −1 −1−1 11 −1−1 −1 11
They are all symmetric. Is this a coincidence?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
What Do You Notice ...
If we look at the matrices we used for the last 3 examples,what do they have in common?
[−2 22 1
] [13 66 −3
] 11 −1 −1−1 11 −1−1 −1 11
They are all symmetric. Is this a coincidence?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
The Spectral Theorem
The Spectral Theorem
A matrix is orthogonally diagonalizable iff it is symmetric.
The proof of this is omitted in the text and is beyond thescope of an undergraduate class. If interested, you can lookat Schur’s Theorem, which deals with unitary matrices andupper triangular matrices with eigenvalues on the maindiagonal. A unitary matrix is a matrix U such that UUT = Ifor real matrices.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
The Spectral Theorem
The Spectral Theorem
A matrix is orthogonally diagonalizable iff it is symmetric.
The proof of this is omitted in the text and is beyond thescope of an undergraduate class. If interested, you can lookat Schur’s Theorem, which deals with unitary matrices andupper triangular matrices with eigenvalues on the maindiagonal. A unitary matrix is a matrix U such that UUT = Ifor real matrices.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Consequences of the Spectral Theorem
I Eigenvalues of symmetric matrices are always real.
I The dimension of an eigenspace for a symmetric matrixis always equal to the multiplicity of the eigenvalue as aroot of the characteristic equation.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Consequences of the Spectral Theorem
I Eigenvalues of symmetric matrices are always real.
I The dimension of an eigenspace for a symmetric matrixis always equal to the multiplicity of the eigenvalue as aroot of the characteristic equation.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
We can think of diagonalization as a way to factor a matrix.But, we have seen that there are conditions for diagonalizinga matrix and even more for orthogonally diagonalizing amatrix. When these fail, we can look to Jordan form for afactorization. But all of these rely on one common fact: thematrix must be square.
What if the matrix is not square? Is there a way to factor it?
If we have an n ×m matrix A with linearly independentcolumns a1, . . . , am, then we can apply the Gram-Schmidtorthonormalization process to these vectors and thisamounts to factoring the matrix. Note that n is notnecessarily equal to m.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
We can think of diagonalization as a way to factor a matrix.But, we have seen that there are conditions for diagonalizinga matrix and even more for orthogonally diagonalizing amatrix. When these fail, we can look to Jordan form for afactorization. But all of these rely on one common fact: thematrix must be square.
What if the matrix is not square? Is there a way to factor it?
If we have an n ×m matrix A with linearly independentcolumns a1, . . . , am, then we can apply the Gram-Schmidtorthonormalization process to these vectors and thisamounts to factoring the matrix. Note that n is notnecessarily equal to m.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
We can think of diagonalization as a way to factor a matrix.But, we have seen that there are conditions for diagonalizinga matrix and even more for orthogonally diagonalizing amatrix. When these fail, we can look to Jordan form for afactorization. But all of these rely on one common fact: thematrix must be square.
What if the matrix is not square? Is there a way to factor it?
If we have an n ×m matrix A with linearly independentcolumns a1, . . . , am, then we can apply the Gram-Schmidtorthonormalization process to these vectors and thisamounts to factoring the matrix. Note that n is notnecessarily equal to m.
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
QR Factorization
QR FactorizationSuppose A = [a1 a2 . . . am] is an n ×m matrix withlinearly independent columns. Then there exists an n ×mmatrix with orthonormal columns Q = [q1 q2 . . . qm] andan m ×m upper triangular matrix R whose diagonal entriesare positive such that A = QR.
So how do we find this Q and R?
The Process
I Apply the Gram-Schmidt process to the columns of Ato find the orthonormal basis {q1,q2, . . . ,qm}
I Take these as the columns of Q
I Solve for R using matrix multiplication in the equationR = QTA
Why are we not using Q−1?
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]
v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 =
s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]−
1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]
=1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
Example
Find the QR factorization of A =
[1 −12 1
].
First, we perform the Gram-Schmidt process on thesevectors.
v1 =
[12
]v2 = s2 − projv1s2
=
[−11
]− 1
5
[12
]=
1
5
[−63
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=
1√5
[12
]=
[1√52√5
]v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]
v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]v2‖v2‖
=
1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1 −6
52 3
5
]?
Why not?
v1‖v1‖
=1√5
[12
]=
[1√52√5
]v2‖v2‖
=1√95
[−6
535
]
=
[− 2√
51√5
]
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1√5− 2√
52√5
1√5
]. Now what?
R = QTA =
[1√5
2√5
− 2√5
1√5
] [1 −12 1
]=
[5√5
1√5
0 3√5
]
So, we have
QR =
[1√5− 2√
52√5
1√5
][√5 1√
5
0 3√5
]=
[1 −12 1
]= A
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1√5− 2√
52√5
1√5
]. Now what?
R = QTA
=
[1√5
2√5
− 2√5
1√5
] [1 −12 1
]=
[5√5
1√5
0 3√5
]
So, we have
QR =
[1√5− 2√
52√5
1√5
][√5 1√
5
0 3√5
]=
[1 −12 1
]= A
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1√5− 2√
52√5
1√5
]. Now what?
R = QTA =
[1√5
2√5
− 2√5
1√5
] [1 −12 1
]
=
[5√5
1√5
0 3√5
]
So, we have
QR =
[1√5− 2√
52√5
1√5
][√5 1√
5
0 3√5
]=
[1 −12 1
]= A
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1√5− 2√
52√5
1√5
]. Now what?
R = QTA =
[1√5
2√5
− 2√5
1√5
] [1 −12 1
]=
[5√5
1√5
0 3√5
]
So, we have
QR =
[1√5− 2√
52√5
1√5
][√5 1√
5
0 3√5
]=
[1 −12 1
]= A
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 1
So, we have Q =
[1√5− 2√
52√5
1√5
]. Now what?
R = QTA =
[1√5
2√5
− 2√5
1√5
] [1 −12 1
]=
[5√5
1√5
0 3√5
]
So, we have
QR =
[1√5− 2√
52√5
1√5
][√5 1√
5
0 3√5
]=
[1 −12 1
]= A
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Example
Find the QR factorization of A =
1 1 01 0 10 1 1
.
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v1 =
110
v2 =
101
− v1 · s2‖v1‖2
110
=
101
− 1
2
110
=
12−1
21
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v3 =
011
− s3 · v1‖v1‖2
110
− s3 · v2‖v2‖2
12−1
21
=
011
− 1
2
110
− 1232
12−1
21
=
011
−1
2120
− 1
6−1
613
=
−23
2323
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v3 =
011
− s3 · v1‖v1‖2
110
− s3 · v2‖v2‖2
12−1
21
=
011
− 1
2
110
− 1232
12−1
21
=
011
−1
2120
− 1
6−1
613
=
−23
2323
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v3 =
011
− s3 · v1‖v1‖2
110
− s3 · v2‖v2‖2
12−1
21
=
011
− 1
2
110
− 1232
12−1
21
=
011
−1
2120
− 1
6−1
613
=
−23
2323
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v3 =
011
− s3 · v1‖v1‖2
110
− s3 · v2‖v2‖2
12−1
21
=
011
− 1
2
110
− 1232
12−1
21
=
011
−1
2120
− 1
6−1
613
=
−23
2323
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
v3 =
011
− s3 · v1‖v1‖2
110
− s3 · v2‖v2‖2
12−1
21
=
011
− 1
2
110
− 1232
12−1
21
=
011
−1
2120
− 1
6−1
613
=
−23
2323
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =
1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
Now, we normalize ...
q1 =1√2
110
=
1√21√2
0
q2 =
1√32
12−1
21
=
1√6
− 1√6
2√6
q3 =
1√43
−23
2323
=
−1√3
1√31√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
So, we have
Q =
1√2
1√6− 1√
31√2− 1√
61√3
0 2√6
1√3
and
R = QTA =
2√2
1√2
1√2
0 3√6
1√6
0 0 2√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 2
So, we have
Q =
1√2
1√6− 1√
31√2− 1√
61√3
0 2√6
1√3
and
R = QTA =
2√2
1√2
1√2
0 3√6
1√6
0 0 2√3
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Example
Find the QR factorization of
A =
1 2 02 3 10 1 −11 4 4
Remember the process:
1. Apply the Gram-Schmidt process to the columns
2. Normalize the vectors
3. Take these vectors as the columns of Q
4. Multiply QTA to find R
1 Introduction
2 OrthogonalMatrices
3 OrthogonallyDiagonalizableMatrices
4 The SpectralTheorem
5 QR Factorization
6 QR Example 1
7 QR Example 2
8 QR Example 3
Example 3
Jumping to the punchline ...
Q =
1√6
0 − 1√6√
23 − 1√
60
0 1√6−√
23
1√6
√23
1√6
R =
√
6 2√
6√
6
0√
6√
6
0 0√
6