This lecture discusses some of the properties of the eigenvalues and eigenvectors of a square matrix.

Table of contents

The first property concerns the eigenvalues of the transpose of a matrix.

Proposition Let be a square matrix. A scalar is an eigenvalue of if and only if it is an eigenvalue of .

Proof

Remember that a scalar is an eigenvalue of if and only if it solves the characteristic equationwhere denotes the determinant. We know that transposition does not change the determinant. Thus,Therefore, is an eigenvalue of if and only ifwhich is verified if and only if is also an eigenvalue of .

Even if and have the same eigenvalues, they do not necessarily have the same eigenvectors.

If is an eigenvector of the transpose, it satisfies

By transposing both sides of the equation, we get

The row vector is called a left eigenvector of .

The diagonal elements of a triangular matrix are equal to its eigenvalues.

Proposition Let be a triangular matrix. Then, each of the diagonal entries of is an eigenvalue of .

Proof

Let be a scalar. Thenis triangular because adding a scalar multiple of the identity matrix to only affects the diagonal entries of . In particular, if is a diagonal entry of , then is a diagonal entry of . Since the determinant of a triangular matrix is equal to the product of its diagonal entries, we have thatSince the eigenvalues of satisfy the characteristic equationwe have that is an eigenvalue of if one of the terms of the above product is equal to zero, that is, if for some .

Eigenvalues allow us to tell whether a matrix is invertible.

Proposition Let be a matrix. Then is invertible if and only if it has no zero eigenvalues.

Proof

We know that is an eigenvalue of if and only if it satisfies the characteristic equationTherefore, is not an eigenvalue of if and only ifwhich happens if and only if is invertible (see the section on the determinant of a singular matrix).

The eigenvalues of the inverse are easy to compute.

Proposition Let be a invertible matrix. Then is an eigenvalue of corresponding to an eigenvector if and only if is an eigenvalue of corresponding to the same eigenvector .

Proof

A scalar is an eigenvalue of corresponding to an eigenvector if and only if Since is invertible, and we can multiply both sides of the equation by , so as to obtainorwhich is true if and only if is an eigenvalue of associated to the eigenvector .

An interesting fact is that complex eigenvalues of real matrices always come in conjugate pairs.

Proposition Let be a matrix having real entries. A complex number is an eigenvalue of corresponding to the eigenvector if and only if its complex conjugate is an eigenvalue corresponding to the conjugate vector .

Proof

A scalar is an eigenvalue of corresponding to an eigenvector if and only if By taking the complex conjugate of both sides of the equation, we obtainSince is real, it is equal to its complex conjugate. Therefore,that is, is an eigenvalue of corresponding to the eigenvector .

If we multiply a matrix by a scalar, then all its eigenvalues are multiplied by the same scalar.

Proposition Let be a matrix and a scalar. If is an eigenvalue of corresponding to the eigenvector , then is an eigenvalue of corresponding to the same eigenvector .

Proof

A scalar is an eigenvalue of corresponding to an eigenvector if and only if If we multiply both sides of the equation by the scalar , we getwhich is true if and only if is an eigenvalue of corresponding to the eigenvector .

Let be a natural number. The -th power of a square matrix is

In other words, the -th power is obtained by performing matrix multiplications of by itself.

It is easy to derive the eigenvalues of from those of .

Proposition Let be a matrix. If is an eigenvalue of corresponding to the eigenvector , then is an eigenvalue of corresponding to the same eigenvector .

Proof

A scalar is an eigenvalue of corresponding to an eigenvector if and only if If we pre-multiply both sides of the equation by , we getIf we again pre-multiply both sides by , we obtainWe can proceed in this manner until we getwhich is true if and only if is an eigenvalue of corresponding to the eigenvector .

Remember that a matrix is said to be Hermitian if and only if it equals its conjugate transpose:

Hermitian matrices have the following nice property.

Proposition Let be a matrix. If is Hermitian, then all its eigenvalues are real (i.e., their complex parts are zero).

Proof

Arbitrarily choose an eigenvalue and one of its associated eigenvectors . By the definition of eigenvector, . Note thatwhere denotes the norm of . If we take the conjugate transpose of both sides of the equation just derived, we obtainwhere we have used the fact that the norm is a real number and, as a consequence, complex conjugation leaves it unaffected. Moreover, we can replace in the last equation with because is Hermitian. Thus, we haveandBut implies that has zero complex part.

If a real matrix is symmetric (i.e., ), then it is also Hermitian (i.e., ) because complex conjugation leaves real numbers unaffected. Therefore, by the previous proposition, all the eigenvalues of a real symmetric matrix are real.

Remember that the trace of a matrix is the sum of its diagonal entries.

Proposition Let be a matrix and its eigenvalues. Then,

Proof

To make this proof as simple as possible, we use the concepts of similarity and Schur decomposition, which we have not yet introduced. You might want to skip this proof now and read it after studying these two concepts. By the Schur decomposition, is unitarily similar to an upper triangular matrix . When two matrices are similar, they have the same trace and the same eigenvalues. Moreover, because is triangular, its diagonal entries are its eigenvalues. Therefore,

The next important result links the determinant of a matrix to its eigenvalues.

Proposition Let be a matrix and its eigenvalues. Then,

Proof

As in the previous proof, we use the concepts of similarity and Schur decomposition. By the Schur decomposition, is unitarily similar to an upper triangular matrix . Two similar matrices have the same determinant and the same eigenvalues. Moreover, because is triangular, its diagonal entries are its eigenvalues and its determinant is equal to the product of its diagonal entries. Therefore,

Below you can find some exercises with explained solutions.

Define

Find the eigenvalues of

Solution

Since is triangular, its eigenvalues are equal to its diagonal entries. Therefore, the eigenvalues of areTransposition does not change the eigenvalues and multiplication by doubles them. Thus, the eigenvalues of areThose of the inverse are and those of are

Please cite as:

Taboga, Marco (2021). "Properties of eigenvalues and eigenvectors", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/properties-of-eigenvalues-and-eigenvectors.

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