Q&A: How to Create a Matrix whose Eigenvalues have Specified Algebraic and Geometric Multiplicity

by Justin Skycak (x.com/justinskycak) on May 04, 2024

Cross-posted from here.

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Question

It’s easy to create a diagonalizable matrix with prescribed eigenvalues: just put those eigenvalues in a diagonal matrix, and then conjugate it by your favorite invertible matrix.

But for each of those eigenvalues, the algebraic multiplicity equals the geometric multiplicity. How do you do create a (non-diagonalizable) matrix whose eigenvalues have specified algebraic and geometric multipicity?

Answer

You can use the same method but with a more general Jordan block matrix instead of a strictly diagonal matrix.

EXAMPLE

Setup

For example, say you want a matrix with an eigenvalue $\lambda_1 = 10$ with algebraic multiplicity $a_1=3$ and geometric multiplicity $g_1=2.$

Just set up your Jordan block matrix like this:

$$\begin{align*} J = \begin{pmatrix} 10 & 1 & 0 \\ 0 & 10 & 0 \\ 0 & 0 & 10 \end{pmatrix} \end{align*}$$

Verifying algebraic multiplicity

The characteristic polynomial of this matrix is $\det(J - \lambda I) = (10 - \lambda)^3,$ so the algebraic multiplicity of the eigenvalue $\lambda_1=10$ is $a_1 = 3$ as desired.

Verifying geometric multiplicity

The geometric multiplicity of the eigenvalue $\lambda_1=10$ is the dimension of this eigenvalue’s eigenspace, that is, the dimension of the solution space ${ v: (J-10I)v = 0 }.$

Writing $v$ as its components $v = \left< v_1, v_2, v_3 \right>,$ the equation $(J-10I)v=0$ becomes

$$\begin{align*} \left< v_2, 0, 0 \right> = \left< 0, 0, 0 \right> \end{align*}$$

which means the solution space consists of vectors $v=\left< v_1, 0, v_3 \right>$ where $v_1$ and $v_3$ are free variables. The two free variables yield a two-dimensional solution space, as desired.

(At this point it’s easy to see intuitively that the off-diagonal $1$ in $J$ forced $v_2=0,$ i.e., each off-diagonal $1$ in a Jordan form matrix decrements the dimension of the corresponding eigenvalue’s eigenspace.)

Conjugating

Again, you can conjugate $J$ by your favorite invertible matrix and retain the desired properties.

CASE OF MULTIPLE DISTINCT EIGENVALUES

If you want to use multiple eigenvalues, just create a Jordan block for each one.

For instance, suppose you want a matrix with

an eigenvalue $\lambda_1$ with algebraic multiplicity $a_1=3$ and geometric multiplicity $g_1=1,$
an eigenvalue $\color{blue}{\lambda_2}$ with algebraic multiplicity $a_1=3$ and geometric multiplicity $g_1=2,$
an eigenvalue $\color{red}{\lambda_3}$ with algebraic multiplicity $a_1=3$ and geometric multiplicity $g_1=3.$

Here’s the corresponding Jordan block matrix:

$$\begin{align*} J = \begin{pmatrix} \lambda_1 & \mathbf 1 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \lambda_1 & \mathbf 1 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \lambda_1 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{blue}{\lambda_2} & \color{blue}{\mathbf{1}} & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{blue}{\lambda_2} & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{blue}{\lambda_2} & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{red}{\lambda_3} & \color{gray} 0 & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{red}{\lambda_3} & \color{gray} 0 \\ \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{gray} 0 & \color{red}{\lambda_3} \end{pmatrix} \end{align*}$$

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