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DIAGONALIZATION: SYMMETRICAND HERMITIAN MATRICES

Symmetric and hermitian matrices, which arise in many applications, enjoy theproperty of always being diagonalizable. Also the set of eigenvectors of such matricescan always be chosen as orthonormal. The diagonalization procedure is essentially thesame as outlined in Sec. 5.3, as we will see in our examples.

Example 1 The horizontal motion of the system of masses and springs where all themasses are the same and the springs are the same, can be analyzed by diagonalizingthe symmetric matrix

Diagonalize .

Solution We have

so that the eigenvalues are and . Eigenvectors are found by solving

; these equations and solutions are

:

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:

Now and are orthogonal since . If we normalize and

, we have the choices for eigenvectors

and forms an orthonormal set. Finally, with

we can write

Since is an orthogonal matrix, . Also represents a rotation of

radians clockwise.

Example 2 Diagonalize

Choose as an orthogonal matrix.

Solution The characteristic equation is , which has solutions 1 and

. The eigenvalue 1 has multiplicity 2, and the eigenvalue is simple (multiplicity

1). Now to determine the eigenvectors, we solve equations.

:

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:

We see that corresponding to the eigenspace is two-dimensional. Therefore,

we can choose a basis for (the basis problem has returned!) of orthonormal

vectors. In fact, choosing first , and second , , we have

and

which are orthogonal. Normalizing ( is already normalized) yields

Not as much work is required for since the eigenspace for this eigenvalue is

one-dimensional. Normalizing , we have

Therefore

and

Diagonalize :

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The last two examples illustrate the basic results for diagonalization of symmetricmatrices.

Theorem 5.4.1 If is symmetric with real entries, then

(a)The eigenvalues are real.

(b)Eigenvectors corresponding to distinct eigenvalues are orthogonal.

(c)The eigenspaces of each eigenvalue have orthogonal bases. The dimension of aneigenspace corresponds to the multiplicity of the eigenvalue.

(d)Matrix is orthogonally diagonalizable; that is, there exists an orthogonalmatrix such that

and so

Proof. We prove only parts (a) and (b). Parts (c) and (d) are proved in more advancedtexts.

(a) Suppose that and that is the corresponding eigenvector.

Therefore,

Carrying out the multiplications and setting real and imaginary parts equal, we find

Now take the dot product of the first equation with and the second with to find

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or

Because and are symmetric and . Thus

and since the transpose of a product is the product of the transposes in reverseorder, we have

Finally, taking the transpose of both sides of the second equation results in

Subtracting the equations yields

or

Because both and cannot be zero (for if so, and could not be an

eigenvector), we must have . Therefore is real.

(b) Let and be eigenvectors corresponding to and , , . We

want to show that . Now

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(By symmetry)

If , then . If , then and we must still have

.

Theorem 5.4.1c tells us that we can find an orthogonal basis for each eigenspace. Thismay require the Gram-Schmidt process, as the next example shows.

Example 3 Orthogonally diagonalize

That is, diagonalize with an orthogonal matrix .

Solution The characteristic polynomial is which has roots

(multiplicity 2) and 2 (simple). To determine eigenvectors, we solve :

:

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:

Since rank , the dimension of is 2.

Looking at and putting , we have

in the eigenspace. Setting , we find

in the eigenspace. Now and are not orthogonal to each other, but they are

linearly independent and span the eigenspace. Using the Gram-Schmidt process on, we find

and

as orthonormal basis vectors for the eigenspace of . Letting ,

we obtain an orthonormal basis (for ) of eigenvectors of . Choosing as

we have

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Now that we can orthogonally diagonalize symmetric matrices, we can consider anapplication to analytic geometry.

Quadratic Forms and Conic Sections A classical problem of analytic geometry is thefollowing:

For the conic section centered at the origin of the plane, described by

(5.4.1)

determine whether the conic section is an ellipse, a hyperbola, or aparabola. Graph the conic section.

A symmetric matrix can be used to describe the left-hand side of Eq. 5.4.1. Inparticular,

Let us call

the matrix of the conic section.5.3Making a change of basis with the orthogonal matrix which diagonalizes , we write

or

Substituting and denoting

by , we have

which reduces to

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or

The last equation is easy to classify and graph in the plane since it has no

``mixed'' term .

Example 4 Classify and graph it.

Solution (We already know the graph since the equation can be rewritten as .

This will make it easy to check our answer.) The matrix of the conic section is

and has eigenpairs

so that

and

Therefore, with the change of variables (basis)

which is the same as

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we have

or

Therefore, the conic section is a hyperbola. To sketch the graph, we must determinethe axes. Since the point gives and

gives , the and axes are as shown in Fig. 5.4.1.

The graph of the hyperbola is also shown in Fig. 5.4.1. Note that the new axes containthe eigenvectors.

Example 5 Classify the conic section

and graph it.

Solution The matrix of the conic section is

Eigenpairs of are

and

Therefore in coordinates defined as in Example 4 (the eigenvectors are the

same) we have

or

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which describes an ellipse. The graph of the ellipse is shown in Fig. 5.4.2. Note thatthe new axes contain the eigenvectors of the matrix. Also note that axes are

obtained by a 45 counterclockwise rotation, which is the action of . Moreover, is defined by the first eignvector, and is defined by the second eigenvector.

Those who have solved these types of conic section problems in calculus realize thatthis linear algebra method of removing the term is much simpler. Of course, a lot

of power machinery had to be developed to get to this point.

In general, the problem of removing the term in is known as the

problem of diagonalizing a quadratic form. This problem arises in many areas;statistics and physics are two. A real quadratic form in the variables is a

function given by

(5.4.2)

where is in . Written out, a real quadratic form in looks like

where through are real numbers. Note that each term has degree 2--hence the

name quadratic form. Our basic theorem about diagonalization of symmetric matricesmeans that any real quadratic form can be diagonalized. So there are new variables

such that in the new variables

This follows from the fact that the matrix in Eq. (5.4.2) can always be chosen as

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symmetric, and symmetric matrices are orthogonally diagonalizable.

Diagonalization in the Hermitian Case Theorem 5.4.1 with a slight change of wordingholds true for hermitian matrices.

If is hermitian, then

The eigenvalues are real.1.Eigenvectors corresponding to distinct eigenvalues are orthogonal.2.The eigenspaces of each eigenvalue have orthogonal bases. The dimension of aneigenspace corresponds to the multiplicity of the eigenvalue.

3.

Matrix is unitarily diagonalizable. That is, there exists a unitary matrix) such that

thus

4.

The proofs of 1 and 2 are almost the same as in Theorem 5.4.1a and b. The differenceis that is used instead of and in , .

Example 6 Can

be unitarily diagonalized? If so, perform the diagonalization.

Solution

Because is hermitian, it can be unitarily diagonalized. Now to find the eigenpairs,

So we have and . Eigenpairs are

To find , we normalize the eigenvectors and use them for the columns of . The

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normalized eigenvectors are found by calculating

and

(Remember that .) Thus we have

and

When a hermitian matrix is diagonalized, the set of orthonormal eigenvectors of is called the set of principal axes of and the associated matrix is called aprincipal axis transformation. For a real hermitian matrix, the principal axistransformation allows us to analyze geometrically.

Example 7 Consider defined by

This can be diagonalized with

so that

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Now represents rotation of counterclockwise while represents rotation of clockwise. If we want to see what does to

we can look at

and we see that is a

Rotation of clockwise1.Stretch of 3 in the first component and a reversal in the second component2.Rotation of counterclockwise3.

We can say even more by determining what does to the unit circle. In the new

coordinates, the unit circle is unchanged because and represent rotations.However, in the new coordinates we have the action of as changing the unit circleby reflecting about the line which is defined by

span

and then transforming the reflected circle to an ellipse, as shown in Fig. 5.4.3.

Finally, we note that in diagonalizing a quadratic form for a conic section, the newaxes obtained from the rotation are exactly the principal axes of the matrix for thequadratic form.

Subsections

PROBLEMS 5.4

Next: PROBLEMS 5.4 Up: Eigenvalues and Diagonalization Previous: PROBLEMS 5.3

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