Producing defective matrix examples

Question

It's easy to come up with examples of interesting diagonalizable matrices $A\in M_n(\mathbb{C})$ with a prescribed set of eigenvalues $\lambda_1,\dots,\lambda_n$ because you can start with your diagonal matrix $\Lambda=\operatorname{diag}(\lambda_1,\dots,\lambda_n)$ and conjugate it by your favorite invertible matrix. Here, let's say "interesting" means non-sparse and not block-diagonal. I'd like to produce lots of examples of defective matrices in a similar way.

In teaching about algebraic and geometric multiplicty, I was trying to come up with nice, larger examples of defective matrices. Is there a similar way to start with a set of distinct eigenvalues $\mu_1,\dots,\mu_k\in\mathbb{C}$, corresponding algebraic multiplicities $a_1,\dots,a_k$ (where $\sum_{i}a_i=n$) and corresponding geometric multiplicities $g_1,\dots,g_k$ (where $1\leq g_i\leq a_i$) and generate lots of examples of interesting matrices with these prescribed properties?

Answer 1

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:

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

Answer 2

Another method to create examples is to begin with an orthonormal basis and build projection matrices which can then be assigned weights as desired. For example, if $\beta = \{ u_1, u_2, \dots , u_n \}$ is an orthonormal basis then $$ A = \lambda_1u_1u_1^T+ \cdots + \lambda_nu_nu_n^T $$ is a matrix with eigenvalues $\lambda_1, \dots , \lambda_n$ for which $\beta$ is an orthonormal eigenbasis. Try it out, since $u_i^Tu_j = \delta_{ij}$ it follows $Au_i = \lambda_i u_i$. Similarly, you can build Jordan forms via reverse engineering with $\beta$, for instance: $$ A = \lambda u_1u_1^T+\lambda u_2u_2^T+ \cdots + \lambda u_nu_n^T+u_1u_2^T+u_2u_3^T+ \cdots + u_{n-1}u_n^T $$ Makes $\beta$ an $n$-chain with eigenvalue $\lambda$. Observe, $$ Au_1 = \lambda u_1, \ \ Au_2 = \lambda u_2+u_1, \ \dots \ , Au_n = \lambda u_n+u_{n-1} $$ Or, $(A-\lambda I)u_1=0$ and $(A-\lambda I)u_i = u_{i-1}$ for $i=2, \dots , n$.

Alternatively, it's sometimes interesting to create examples via the connection to differential equations. For instance, I know the solution of $(D^2+1)^2[y]=0$ or $y''''+2y''+y=0$ includes $\cos t, \sin t$ as well as curious $t\cos t$ and $t \sin t$ solutions.Thus, if I convert this 4th order ODE to a corresponding 4th order linear system of first order ODEs then I know the solution will also require $t\cos t$ type terms. But, that means the matrix in question must have $\lambda = \pm i$ with an algebraic multiplicity of two, yet a geometric multiplicity of one. Reduction of order is the method here: $$ x_1 = y$$ $$ x_2 = y' \ \ \Rightarrow x_1' = x_2 $$ $$ x_3 = y'' \ \ \Rightarrow x_2' = x_3 $$ $$ x_4 = y''' \ \ \Rightarrow x_3' = x_4 $$ Thus $x_4' = y'''' = -2y''-y = -2x_3-x_1$ and the system is thus: $$ \frac{dx}{dt} = Ax \qquad \text{where} \qquad A = \left[ \begin{array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -1 & 0 & -2 & 0 \end{array} \right] $$ This makes a relatively friendly test question which isn't just directly a Jordan form. But, to be honest, even giving Jordan form and asking for direct interpretation from the pattern is challenging for many students. But, it's worth it because the Jordan form is totally cool.