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Properties of eigenvalues and eigenvectors

by , PhD

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

Table of Contents

Left eigenvectors

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

Proposition Let A be a $K	imes K$ square matrix. A scalar $lambda $ is an eigenvalue of A if and only if it is an eigenvalue of $A^{intercal }$.

Proof

Remember that a scalar $lambda $ is an eigenvalue of A if and only if it solves the characteristic equation[eq1]where $det $ denotes the determinant. We know that transposition does not change the determinant. Thus,[eq2]Therefore, $lambda $ is an eigenvalue of A if and only if[eq3]which is verified if and only if $lambda $ is also an eigenvalue of $A^{intercal }$.

Even if A and $A^{intercal }$ have the same eigenvalues, they do not necessarily have the same eigenvectors.

If $y$ is an eigenvector of the transpose, it satisfies[eq4]

By transposing both sides of the equation, we get[eq5]

The row vector $y^{intercal }$ is called a left eigenvector of A.

Eigenvalues of a triangular matrix

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

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

Proof

Let $lambda $ be a scalar. Then[eq6]is triangular because subtracting a scalar multiple of the identity matrix from A only affects the diagonal entries of A. In particular, if $A_{kk}$ is a diagonal entry of A, then $A_{kk}-lambda $ is a diagonal entry of $A-lambda I$. Since the determinant of a triangular matrix is equal to the product of its diagonal entries, we have that[eq7] Since the eigenvalues of A satisfy the characteristic equation[eq8]we have that $lambda $ is an eigenvalue of A if one of the terms [eq9] of the above product is equal to zero, that is, if $lambda =A_{kk}$ for some k.

Zero eigenvalues and invertibility

Eigenvalues allow us to tell whether a matrix is invertible.

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

Proof

We know that $lambda $ is an eigenvalue of A if and only if it satisfies the characteristic equation[eq10]Therefore, 0 is not an eigenvalue of A if and only if[eq11]which happens if and only if A is invertible (see the section on the determinant of a singular matrix).

Eigenvalues and eigenvectors of the inverse matrix

The eigenvalues of the inverse are easy to compute.

Proposition Let A be a $K	imes K$ invertible matrix. Then $lambda $ is an eigenvalue of A corresponding to an eigenvector x if and only if $lambda ^{-1}$ is an eigenvalue of $A^{-1}$ corresponding to the same eigenvector x.

Proof

A scalar $lambda $ is an eigenvalue of A corresponding to an eigenvector x if and only if [eq12]Since A is invertible, $lambda 
eq 0$ and we can multiply both sides of the equation by [eq13], so as to obtain[eq14]or[eq15]which is true if and only if $lambda ^{-1}$ is an eigenvalue of $A^{-1}$ associated to the eigenvector x.

Conjugate pairs

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

Proposition Let A be a $K	imes K$ matrix having real entries. A complex number $lambda $ is an eigenvalue of A corresponding to the eigenvector x if and only if its complex conjugate [eq16] is an eigenvalue corresponding to the conjugate vector $overline{x}$.

Proof

A scalar $lambda $ is an eigenvalue of A corresponding to an eigenvector x if and only if [eq17]By taking the complex conjugate of both sides of the equation, we obtain[eq18]Since A is real, it is equal to its complex conjugate. Therefore,[eq19]that is, [eq20] is an eigenvalue of A corresponding to the eigenvector $overline{x}$.

Scalar multiples

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

Proposition Let A be a $K	imes K$ matrix and $lpha 
eq 0$ a scalar. If $lambda $ is an eigenvalue of A corresponding to the eigenvector x, then $lpha lambda $ is an eigenvalue of $lpha A$ corresponding to the same eigenvector x.

Proof

A scalar $lambda $ is an eigenvalue of A corresponding to an eigenvector x if and only if [eq21]If we multiply both sides of the equation by the scalar $lpha $, we get[eq22]which is true if and only if $lpha lambda $ is an eigenvalue of $lpha A$ corresponding to the eigenvector x.

Matrix powers

Let n be a natural number. The n-th power of a square matrix A is[eq23]

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

It is easy to derive the eigenvalues of $A^{n}$ from those of A.

Proposition Let A be a $K	imes K$ matrix. If $lambda $ is an eigenvalue of A corresponding to the eigenvector x, then $lambda ^{n}$ is an eigenvalue of $A^{n}$ corresponding to the same eigenvector x.

Proof

A scalar $lambda $ is an eigenvalue of A corresponding to an eigenvector x if and only if [eq24]If we pre-multiply both sides of the equation by A, we get[eq25]If we again pre-multiply both sides by A, we obtain[eq26]We can proceed in this manner until we get[eq27]which is true if and only if $lambda ^{n}$ is an eigenvalue of $A^{n}$ corresponding to the eigenvector x.

All the eigenvalues of a Hermitian matrix are real

Remember that a matrix A is said to be Hermitian if and only if it equals its conjugate transpose:[eq28]

Hermitian matrices have the following nice property.

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

Proof

Arbitrarily choose an eigenvalue $lambda $ and one of its associated eigenvectors x. By the definition of eigenvector, $x
eq 0$. Note that[eq29]where [eq30] denotes the norm of x. If we take the conjugate transpose of both sides of the equation just derived, we obtain[eq31]where 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 $A^{st }$ in the last equation with A because A is Hermitian. Thus, we have[eq32]and[eq33]But [eq34] implies that $lambda $ has zero complex part.

All the eigenvalues of a symmetric real matrix are real

If a real matrix A is symmetric (i.e., $A=A^{	op }$), then it is also Hermitian (i.e., [eq35]) because complex conjugation leaves real numbers unaffected. Therefore, by the previous proposition, all the eigenvalues of a real symmetric matrix are real.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Define[eq36]

Find the eigenvalues of [eq37]

Solution

Since A is triangular, its eigenvalues are equal to its diagonal entries. Therefore, the eigenvalues of A are[eq38]Transposition does not change the eigenvalues and multiplication by $2$ doubles them. Thus, the eigenvalues of [eq39] are[eq40]Those of the inverse [eq41] are [eq42]and those of [eq43] are[eq44]

How to cite

Please cite as:

Taboga, Marco (2017). "Properties of eigenvalues and eigenvectors", Lectures on matrix algebra. https://www.statlect.com/matrix-algebra/properties-of-eigenvalues-and-eigenvectors.

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