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Algebraic and geometric multiplicity of eigenvalues

by , PhD

The algebraic multiplicity of an eigenvalue is the number of times it appears as a root of the characteristic polynomial (i.e., the polynomial whose roots are the eigenvalues of a matrix).

The geometric multiplicity of an eigenvalue is the dimension of the linear space of its associated eigenvectors (i.e., its eigenspace).

In this lecture we provide rigorous definitions of the two concepts of algebraic and geometric multiplicity and we prove some useful facts about them.

Table of Contents

Algebraic multiplicity

Let us start with a definition.

Definition Let A be a $K	imes K$ matrix. Denote by [eq1] the K possibly repeated eigenvalues of A, which solve the characteristic equation[eq2]We say that an eigenvalue $lambda _{k}$ has algebraic multiplicity [eq3] if and only if there are no more and no less than [eq4] solutions of the characteristic equation equal to $lambda _{k} $.

Let us see some examples.

Example Consider the $2	imes 2$ matrix [eq5]The characteristic polynomial is[eq6]The roots of the polynomial, that is, the solutions of [eq7] are[eq8]Thus, A has two distinct eigenvalues. Their algebraic multiplicities are [eq9]because they are not repeated.

Example Define the $2	imes 2$ matrix [eq10]Its characteristic polynomial is[eq11]The roots of the polynomial, that is, the solutions of [eq7] are[eq13]Thus, A has one repeated eigenvalue whose algebraic multiplicity is [eq14]

Geometric multiplicity

Recall that each eigenvalue is associated to a linear space of eigenvectors, called eigenspace.

Definition Let A be a $K	imes K$ matrix. Let $lambda _{k}$ be one of the eigenvalues of A and denote its associated eigenspace by $E_{k}$. The dimension of $E_{k}$ is called the geometric multiplicity of the eigenvalue $lambda _{k}$.

Let's now make some examples.

Definition Consider the $2	imes 2$ matrix [eq15]The characteristic polynomial is[eq16]The roots of the polynomial are[eq17]The eigenvectors associated to $lambda _{1}=2$ are the vectors [eq18]that solve the equation[eq19]or[eq20]The last equation implies that[eq21]Therefore, the eigenspace of $lambda _{1}$ is the linear space that contains all vectors $x_{1}$ of the form[eq22]where $lpha $ can be any scalar. Thus, the eigenspace of $lambda _{1}$ is generated by a single vector[eq23]Therefore, it has dimension 1. As a consequence, the geometric multiplicity of $lambda _{1}$ is 1.

Example Consider the $2	imes 2$ matrix [eq24]The characteristic polynomial is[eq25]and its roots are[eq13]Thus, there is a repeated eigenvalue ([eq27]) with algebraic multiplicity equal to 2. Its associated eigenvectors [eq28]solve the equation[eq29]or[eq30]The equation is satisfied for $x_{11}=0$ and any value of $x_{21}$. As a consequence, the eigenspace of $lambda _{1}$ is the linear space that contains all vectors $x_{1}$ of the form[eq31]where $lpha $ can be any scalar. Since the eigenspace of $lambda _{1}$ is generated by a single vector[eq32]it has dimension 1. As a consequence, the geometric multiplicity of $lambda _{1}$ is 1, less than its algebraic multiplicity, which is equal to 2.

Example Define the $2	imes 2$ matrix [eq33]The characteristic polynomial is[eq34]and its roots are[eq35]Thus, there is a repeated eigenvalue ([eq36]) with algebraic multiplicity equal to 2. Its associated eigenvectors [eq28]solve the equation[eq38]or[eq39]The equation is satisfied for any value of $x_{11}$ and $x_{21}$. As a consequence, the eigenspace of $lambda _{1}$ is the linear space that contains all vectors $x_{1}$ of the form[eq40]where $x_{11}$ and $x_{21}$ are scalars that can be arbitrarily chosen. Thus, the eigenspace of $lambda _{1}$ is generated by the two linearly independent vectors[eq41]Hence, it has dimension $2$. As a consequence, the geometric multiplicity of $lambda _{2}$ is 2, equal to its algebraic multiplicity.

A takeaway message from the previous examples is that the algebraic and geometric multiplicity of an eigenvalue do not necessarily coincide.

Relationship between algebraic and geometric multiplicity

The following proposition states an important property of multiplicities.

Proposition Let A be a $K	imes K$ matrix. Let $lambda _{k}$ be one of the eigenvalues of A. Then, the geometric multiplicity of $lambda _{k}$ is less than or equal to its algebraic multiplicity.

Proof

Suppose that the geometric multiplicity of $lambda _{k}$ is equal to $r$, so that there are $r$ linearly independent eigenvectors [eq42] associated to $lambda _{k}$. Arbitrarily choose $K-r$ vectors [eq43], all having dimension Kx1 and such that the $K $ column vectors [eq44] are linearly independent. Define the $K	imes K$ matrix [eq45]For any $r$, denote by $b_{r}$ the vector that solves[eq46]which is guaranteed to exist because X is full-rank (its columns are linearly independent). Define the [eq47] matrix[eq48]and denote by $C$ its upper [eq49] block and by $D$ its lower [eq50] block:[eq51]Denote by [eq52] the $K	imes K$ identity matrix. For any scalar $lambda $, we have that[eq53]Since X is full-rank and, as a consequence its determinant is non-zero, we can write[eq54]where in step $rame{A}$ we have used a result about the determinant of block-matrices. The eigenvalues of A solve the characteristic equation [eq55]or, equivalently, the equation[eq56]This equation has a root [eq57] that is repeated at least $r $ times. Therefore, the algebraic multiplicity of $lambda _{k}$ is at least equal to its geometric multiplicity $r$. It can be larger if $lambda _{k}$ is also a root of [eq58]

Defective eigenvalues

When the geometric multiplicity of a repeated eigenvalue is strictly less than its algebraic multiplicity, then that eigenvalue is said to be defective.

An eigenvalue that is not repeated has an associated eigenvector which is different from zero. Therefore, the dimension of its eigenspace is equal to 1, its geometric multiplicity is equal to 1 and equals its algebraic multiplicity. Thus, an eigenvalue that is not repeated is also non-defective.

Solved exercises

Below you can find some exercises with explained solutions.

Exercise 1

Find whether the matrix[eq59]has any defective eigenvalues.

Solution

The characteristic polynomial is[eq60]and its roots are[eq61]Thus, there are no repeated eigenvalues and, as a consequence, no defective eigenvalues.

Exercise 2

Define[eq62]

Determine whether A possesses any defective eigenvalues.

Solution

The characteristic polynomial is[eq63]where in step $rame{A}$ we have used the Laplace expansion along the third row. The roots of the polynomial are[eq64]Thus, there is a repeated eigenvalue ([eq65]) with algebraic multiplicity equal to 2. Its associated eigenvectors [eq66]solve the equation[eq67]or[eq68]The equation is satisfied for any value of $x_{22}$ and $x_{12}=x_{32}=0$. As a consequence, the eigenspace of $lambda _{2}$ is the linear space that contains all vectors $x_{2}$ of the form[eq69]where the scalar $x_{22}$ can be arbitrarily chosen. Therefore, the eigenspace of $lambda _{2}$ is generated by a single vector[eq70]Thus, it has dimension 1, the geometric multiplicity of $lambda _{2}$ is 1, its algebraic multiplicity is 2 and it is defective.

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