Chapter 6 Eigenvalues and Eigenvectors - 6.1 Eigenvalues...
Transcript of Chapter 6 Eigenvalues and Eigenvectors - 6.1 Eigenvalues...
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Chapter 6 Eigenvalues and Eigenvectors6.1 Eigenvalues and Eigenvectors
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
What are eigenvalues and eigenvalues?
DefinitionGiven an n × n matrix A, a nonzero vector u in Rn is calledan eigenvector of A if there exists a scalar λ such thatAu = λu. The scalar λ is called an eigenvalue of A.
Note: λ is allowed to be 0.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
What are eigenvalues and eigenvalues?
DefinitionGiven an n × n matrix A, a nonzero vector u in Rn is calledan eigenvector of A if there exists a scalar λ such thatAu = λu. The scalar λ is called an eigenvalue of A.
Note: λ is allowed to be 0.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940
Does anyone know what the natural frequency of an objectlike a bridge is?
DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.
The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.
Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940
Does anyone know what the natural frequency of an objectlike a bridge is?
DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.
The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.
Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940
Does anyone know what the natural frequency of an objectlike a bridge is?
DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.
The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.
Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940
Does anyone know what the natural frequency of an objectlike a bridge is?
DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.
The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.
Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?Tacoma Narrows Bridge, 1940
Does anyone know what the natural frequency of an objectlike a bridge is?
DefinitionThe natural frequency is the frequency at which a systemnaturally vibrates once it has been set into motion. That is,it is motion a structure takes on in response to wind or beingwalked on.
The reason this bridge collapsed is because the naturalfrequency of the bridge was too close to the naturalfrequency of the wind. When frequencies match, thecompound provided too strong a force for the bridge.
Mathematically, the natural frequency can be characterizedby the eigenvalue of smallest magnitude.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Linear transformations: Eigenvectors make understandinglinear transformations easy. They are the “axes” (directions)along which a linear transformation acts simply bystretching/compressing and/or reflecting.
Eigenvalues give you the factors by which this compressionoccurs.
The more directions you have along which you understandthe behavior of a linear transformation, the easier it is tounderstand the linear transformation; so you want to have asmany linearly independent eigenvectors as possibleassociated to a single linear transformation.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Linear transformations: Eigenvectors make understandinglinear transformations easy. They are the “axes” (directions)along which a linear transformation acts simply bystretching/compressing and/or reflecting.
Eigenvalues give you the factors by which this compressionoccurs.
The more directions you have along which you understandthe behavior of a linear transformation, the easier it is tounderstand the linear transformation; so you want to have asmany linearly independent eigenvectors as possibleassociated to a single linear transformation.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Linear transformations: Eigenvectors make understandinglinear transformations easy. They are the “axes” (directions)along which a linear transformation acts simply bystretching/compressing and/or reflecting.
Eigenvalues give you the factors by which this compressionoccurs.
The more directions you have along which you understandthe behavior of a linear transformation, the easier it is tounderstand the linear transformation; so you want to have asmany linearly independent eigenvectors as possibleassociated to a single linear transformation.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Why are eigenvalues and eigenvectors important?
Eigenvalues tell us a whole lot about a matrix:
1. If A is singular then λ = 0
2. If A is symmetric, then all eigenvalues are real
3. If A has rank n then the eigenvectors form a basis for Rn
4. The eigenvectors of AAT form a basis for col(A)
5. The eigenvectors of ATA form a basis for row(A)
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How these work
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
[3 11 3
],u =
[11
]
[3 11 3
] [11
]=
[44
]= 4
[11
]
Therefore, we have shown u is an eigenvector of A andλ = 4 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How these work
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
[3 11 3
],u =
[11
]
[3 11 3
] [11
]=
[44
]
= 4
[11
]
Therefore, we have shown u is an eigenvector of A andλ = 4 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How these work
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
[3 11 3
],u =
[11
]
[3 11 3
] [11
]=
[44
]= 4
[11
]
Therefore, we have shown u is an eigenvector of A andλ = 4 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How these work
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
[3 11 3
],u =
[11
]
[3 11 3
] [11
]=
[44
]= 4
[11
]
Therefore, we have shown u is an eigenvector of A andλ = 4 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Visualization
u
Au
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Visualization
u
Au
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
10 12 −68 11 −4
34 44 −19
,u =
012
10 12 −68 11 −4
34 44 −19
012
=
036
= 3
012
Therefore, we have shown u is an eigenvector of A andλ = 3 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
10 12 −68 11 −4
34 44 −19
,u =
012
10 12 −68 11 −4
34 44 −19
012
=
036
= 3
012
Therefore, we have shown u is an eigenvector of A andλ = 3 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
10 12 −68 11 −4
34 44 −19
,u =
012
10 12 −68 11 −4
34 44 −19
012
=
036
= 3
012
Therefore, we have shown u is an eigenvector of A andλ = 3 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Verify that u is an eigenvector for the matrix A anddetermine the associated eigenvalue.
A =
10 12 −68 11 −4
34 44 −19
,u =
012
10 12 −68 11 −4
34 44 −19
012
=
036
= 3
012
Therefore, we have shown u is an eigenvector of A andλ = 3 is the associated eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣
= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[3 11 3
]
The characteristic polynomial is found by findingdet(A− λIn), so in this case we need det(A− λI2),
det(A− λI2) =
∣∣∣∣ 3 − λ 11 3 − λ
∣∣∣∣= (3 − λ)2 − 1
= 9 − 6λ+ λ2 − 1
= λ2 − 6λ+ 8
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
How we find eigenvalues
So, our characteristic polynomial is λ2 − 6λ+ 8.
The eigenvalues are the values of λ such thatdet(A− λIn) = 0.
So, here we have
λ2 − 6λ+ 8 = 0
(λ− 2)(λ− 4) = 0
λ = 2, 4
So, the eigenvalues associated with
[3 11 3
]are λ = 2 and
λ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[−7 10−5 8
]
det(A− λI2) =
∣∣∣∣ −7 − λ 10-5 8 − λ
∣∣∣∣= (−7 − λ)(8 − λ) + 50
= −56 − λ+ λ2 + 50
= λ2 − λ− 6
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[−7 10−5 8
]
det(A− λI2) =
∣∣∣∣ −7 − λ 10-5 8 − λ
∣∣∣∣
= (−7 − λ)(8 − λ) + 50
= −56 − λ+ λ2 + 50
= λ2 − λ− 6
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[−7 10−5 8
]
det(A− λI2) =
∣∣∣∣ −7 − λ 10-5 8 − λ
∣∣∣∣= (−7 − λ)(8 − λ) + 50
= −56 − λ+ λ2 + 50
= λ2 − λ− 6
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[−7 10−5 8
]
det(A− λI2) =
∣∣∣∣ −7 − λ 10-5 8 − λ
∣∣∣∣= (−7 − λ)(8 − λ) + 50
= −56 − λ+ λ2 + 50
= λ2 − λ− 6
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Example
Find the characteristic polynomial and eigenvalues for A.
A =
[−7 10−5 8
]
det(A− λI2) =
∣∣∣∣ −7 − λ 10-5 8 − λ
∣∣∣∣= (−7 − λ)(8 − λ) + 50
= −56 − λ+ λ2 + 50
= λ2 − λ− 6
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Now,
λ2 − λ− 6 = 0
(λ− 3)(λ+ 2) = 0
λ = 3,−2
Therefore, the eigenvalues associated with
[−7 10−5 8
]are
λ = 3 and λ = −2.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Now,
λ2 − λ− 6 = 0
(λ− 3)(λ+ 2) = 0
λ = 3,−2
Therefore, the eigenvalues associated with
[−7 10−5 8
]are
λ = 3 and λ = −2.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Now,
λ2 − λ− 6 = 0
(λ− 3)(λ+ 2) = 0
λ = 3,−2
Therefore, the eigenvalues associated with
[−7 10−5 8
]are
λ = 3 and λ = −2.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Another Example
Now,
λ2 − λ− 6 = 0
(λ− 3)(λ+ 2) = 0
λ = 3,−2
Therefore, the eigenvalues associated with
[−7 10−5 8
]are
λ = 3 and λ = −2.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
Example
Find the characteristic polynomial and eigenvalues for A.
A =
1 0 −20 1 22 2 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
det(A− λI3) =
∣∣∣∣∣∣1 − λ 0 -2
0 1 − λ 22 2 1 − λ
∣∣∣∣∣∣
= (1 − λ)
∣∣∣∣ 1 − λ 22 1 − λ
∣∣∣∣− 2
∣∣∣∣ 0 1 − λ2 2
∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))
= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)
= −λ3 + 3λ2 − 3λ+ 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
det(A− λI3) =
∣∣∣∣∣∣1 − λ 0 -2
0 1 − λ 22 2 1 − λ
∣∣∣∣∣∣= (1 − λ)
∣∣∣∣ 1 − λ 22 1 − λ
∣∣∣∣− 2
∣∣∣∣ 0 1 − λ2 2
∣∣∣∣
= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))
= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)
= −λ3 + 3λ2 − 3λ+ 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
det(A− λI3) =
∣∣∣∣∣∣1 − λ 0 -2
0 1 − λ 22 2 1 − λ
∣∣∣∣∣∣= (1 − λ)
∣∣∣∣ 1 − λ 22 1 − λ
∣∣∣∣− 2
∣∣∣∣ 0 1 − λ2 2
∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))
= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)
= −λ3 + 3λ2 − 3λ+ 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
det(A− λI3) =
∣∣∣∣∣∣1 − λ 0 -2
0 1 − λ 22 2 1 − λ
∣∣∣∣∣∣= (1 − λ)
∣∣∣∣ 1 − λ 22 1 − λ
∣∣∣∣− 2
∣∣∣∣ 0 1 − λ2 2
∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))
= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)
= −λ3 + 3λ2 − 3λ+ 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
det(A− λI3) =
∣∣∣∣∣∣1 − λ 0 -2
0 1 − λ 22 2 1 − λ
∣∣∣∣∣∣= (1 − λ)
∣∣∣∣ 1 − λ 22 1 − λ
∣∣∣∣− 2
∣∣∣∣ 0 1 − λ2 2
∣∣∣∣= (1 − λ)((1 − λ)2 − 4) − 2(0 − 2(1 − λ))
= (1 − λ)(λ2 − 2λ+ 3) + 4(1 − λ)
= −λ3 + 3λ2 − 3λ+ 1
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
Now,
−λ3 + 3λ2 − 3λ+ 1 = 0
(1 − λ)3 = 0
λ = 1
So, the only eigenvalue associated with
1 −0 −20 1 22 2 1
is
λ = 1.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
Now,
−λ3 + 3λ2 − 3λ+ 1 = 0
(1 − λ)3 = 0
λ = 1
So, the only eigenvalue associated with
1 −0 −20 1 22 2 1
is
λ = 1.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
Now,
−λ3 + 3λ2 − 3λ+ 1 = 0
(1 − λ)3 = 0
λ = 1
So, the only eigenvalue associated with
1 −0 −20 1 22 2 1
is
λ = 1.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
One More
Now,
−λ3 + 3λ2 − 3λ+ 1 = 0
(1 − λ)3 = 0
λ = 1
So, the only eigenvalue associated with
1 −0 −20 1 22 2 1
is
λ = 1.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the eigenvectors
DefinitionIf λ is an eigenvalue for an n× n matrix A, the eigenspace ofλ is the solution set to the homogeneous linear system(A− λIn)x = 0, that is to say, the subspace ker(A− λIn).
So, to find the vectors we use the matrix A− λIn and rowreduce to find the form of the infinite number of solutions.The vector can be represented as a basis for the eigenspaceassociated with the given eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the eigenvectors
DefinitionIf λ is an eigenvalue for an n× n matrix A, the eigenspace ofλ is the solution set to the homogeneous linear system(A− λIn)x = 0, that is to say, the subspace ker(A− λIn).
So, to find the vectors we use the matrix A− λIn and rowreduce to find the form of the infinite number of solutions.The vector can be represented as a basis for the eigenspaceassociated with the given eigenvalue.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspaces
Example
Find a basis for the eigenspace of A associated with eacheigenvalue.
A =
[3 11 3
]
From earlier, we found that the eigenvalues were λ = 2 andλ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspaces
Example
Find a basis for the eigenspace of A associated with eacheigenvalue.
A =
[3 11 3
]
From earlier, we found that the eigenvalues were λ = 2 andλ = 4.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...
A− 2I2 =
[3 − λ 1
1 3 − λ
]
=
[1 11 1
]∼[
1 10 0
]
Solving this system, we get x = s
[1−1
]So, a basis for the eigenspace associated with λ = 2 is
sp
{[1−1
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...
A− 2I2 =
[3 − λ 1
1 3 − λ
]=
[1 11 1
]
∼[
1 10 0
]
Solving this system, we get x = s
[1−1
]So, a basis for the eigenspace associated with λ = 2 is
sp
{[1−1
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...
A− 2I2 =
[3 − λ 1
1 3 − λ
]=
[1 11 1
]∼[
1 10 0
]
Solving this system, we get x = s
[1−1
]So, a basis for the eigenspace associated with λ = 2 is
sp
{[1−1
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...
A− 2I2 =
[3 − λ 1
1 3 − λ
]=
[1 11 1
]∼[
1 10 0
]
Solving this system, we get x = s
[1−1
]
So, a basis for the eigenspace associated with λ = 2 is
sp
{[1−1
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Finding the bases for the eigenspacesλ = 2: Remember, we are solving the homogeneous system...
A− 2I2 =
[3 − λ 1
1 3 − λ
]=
[1 11 1
]∼[
1 10 0
]
Solving this system, we get x = s
[1−1
]So, a basis for the eigenspace associated with λ = 2 is
sp
{[1−1
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Find the bases for the eigenspaces
λ = 4:
A− 4I2 =
[3 − λ 1
1 3 − λ
]
=
[−1 11 −1
]∼[
1 −10 0
]
Solving this system, we get x = s
[11
]So, a basis for the eigenspace associated with λ = 4 is
sp
{[11
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Find the bases for the eigenspaces
λ = 4:
A− 4I2 =
[3 − λ 1
1 3 − λ
]=
[−1 11 −1
]
∼[
1 −10 0
]
Solving this system, we get x = s
[11
]So, a basis for the eigenspace associated with λ = 4 is
sp
{[11
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Find the bases for the eigenspaces
λ = 4:
A− 4I2 =
[3 − λ 1
1 3 − λ
]=
[−1 11 −1
]∼[
1 −10 0
]
Solving this system, we get x = s
[11
]So, a basis for the eigenspace associated with λ = 4 is
sp
{[11
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Find the bases for the eigenspaces
λ = 4:
A− 4I2 =
[3 − λ 1
1 3 − λ
]=
[−1 11 −1
]∼[
1 −10 0
]
Solving this system, we get x = s
[11
]
So, a basis for the eigenspace associated with λ = 4 is
sp
{[11
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
Find the bases for the eigenspaces
λ = 4:
A− 4I2 =
[3 − λ 1
1 3 − λ
]=
[−1 11 −1
]∼[
1 −10 0
]
Solving this system, we get x = s
[11
]So, a basis for the eigenspace associated with λ = 4 is
sp
{[11
]}.
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
The Big Theorem - Version 8
TheoremLet A = {a1, . . . , an} be a set of n vectors in Rn, letA = [a1 a2 . . . an] and let T : Rn → Rn be given by T (x) = Ax.Then the following are equivalent:
(a) A spans Rn
(b) A is linearly independent
(c) Ax = b has a unique solution ∀b ∈ Rn
(d) T is onto
(e) T is 1-1
(f) A is invertible
(g) ker(T ) = {0}(h) A is a basis for Rn
(i) col(A) = Rn
(j) row(A) = Rn
(k) rank(A) = n
(l) det(A) 6= 0
(m) λ = 0 is not an eigenvalue of A
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
The Big Theorem - Version 8
TheoremLet A = {a1, . . . , an} be a set of n vectors in Rn, letA = [a1 a2 . . . an] and let T : Rn → Rn be given by T (x) = Ax.Then the following are equivalent:
(a) A spans Rn
(b) A is linearly independent
(c) Ax = b has a unique solution ∀b ∈ Rn
(d) T is onto
(e) T is 1-1
(f) A is invertible
(g) ker(T ) = {0}(h) A is a basis for Rn
(i) col(A) = Rn
(j) row(A) = Rn
(k) rank(A) = n
(l) det(A) 6= 0
(m) λ = 0 is not an eigenvalue of A
1 Introduction
2 EigenvectorsAlgebraically
3 FindingEigenvalues
4 FindingEigenspaces
5 Putting itTogether
The Big Theorem - Version 8
TheoremLet A = {a1, . . . , an} be a set of n vectors in Rn, letA = [a1 a2 . . . an] and let T : Rn → Rn be given by T (x) = Ax.Then the following are equivalent:
(a) A spans Rn
(b) A is linearly independent
(c) Ax = b has a unique solution ∀b ∈ Rn
(d) T is onto
(e) T is 1-1
(f) A is invertible
(g) ker(T ) = {0}(h) A is a basis for Rn
(i) col(A) = Rn
(j) row(A) = Rn
(k) rank(A) = n
(l) det(A) 6= 0
(m) λ = 0 is not an eigenvalue of A