Eigenvalues of the linear map $X\mapsto BX+XB$

Question

Let $V = M_{2,2}(\mathbb{C})$, the vector space of all $2 × 2$-matrices over $\mathbb{C}$. Consider the matrix $$B= \left[ \begin{array}{ c c } 0 & 2 \\ 0 & 0 \end{array} \right] $$ Let $f : V → V$ be the function given by $f(X) = BX + XB$ for all $X ∈ V$. Determine the eigenvalues of the map $f$ and find the dimension of each of its non-zero eigenspaces. Determine whether or not $f$ is diagonalisable.

I've been trying this question for a while and have no idea what to do, can someone please explain how do this step by step. Thank you.

Answer 1

It's clear that $f$ is a linear transformation between finite dimensional vector spaces. In particular, it's not hard to see that the matrix representation of $[f]$ is a $4\times 4$ matrix. If we choose the basis \begin{align} \mathcal{B}=\left\{ e_1= \begin{pmatrix} 1 & 0\\ 0& 0 \end{pmatrix}, e_2= \begin{pmatrix} 0 & 1\\ 0& 0 \end{pmatrix}, e_3= \begin{pmatrix} 0 & 0\\ 1& 0 \end{pmatrix}, e_4= \begin{pmatrix} 0 & 0\\ 0& 1 \end{pmatrix} \right\} \end{align} then we see that $[f]_{\mathcal{B}\rightarrow \mathcal{B}}$ is given by \begin{align} [f]_{\mathcal{B}\rightarrow \mathcal{B}} = ([f(e_1)]_\mathcal{B}\ \ [f(e_2)]_\mathcal{B}\ \ [f(e_3)]_\mathcal{B}\ \ [f(e_4)]_\mathcal{B}). \end{align}

Observe \begin{align} f(e_1) =& \begin{pmatrix} 0 & 2\\ 0 & 0 \end{pmatrix} = 2e_2,\ \ f(e_2) = \begin{pmatrix} 0 & 0\\ 0 & 0 \end{pmatrix} \\ f(e_3) =& \begin{pmatrix} 2 & 0\\ 0 & 2 \end{pmatrix} = 2e_1+2e_4,\ \ f(e_4) = \begin{pmatrix} 0 & 2\\ 0 & 0 \end{pmatrix} = 2e_2 \end{align} then we have \begin{align} [f]_{\mathcal{B}\rightarrow \mathcal{B}} = \begin{pmatrix} 0 & 0& 2& 0\\ 2 & 0& 0& 2\\ 0 & 0& 0& 0\\ 0 & 0& 2& 0 \end{pmatrix}. \end{align}

I will leave the rest to the reader.

Answer 2

Do you know what "eigenvalue" means? The direct calculation is pretty straight forward:

Let $X= \begin{bmatrix}a & b \\ c & d \end{bmatrix}$ Then $f(x)= BX+ XB= \begin{bmatrix}0 & 2 \\ 0 & 0 \end{bmatrix}\begin{bmatrix}a & b \\ c & d \end{bmatrix}+ \begin{bmatrix}a & b \\ c & d \end{bmatrix}\begin{bmatrix}0 & 2 \\ 0 & 0 \end{bmatrix}= \begin{bmatrix}2c & 2a+ 2d \\ 0 & 2c\end{bmatrix}$.

A number, $\lambda$, is an Eigenvalue for F if and only if $\begin{bmatrix}2c & 2a+ 2d \\ 0 & 2c\end{bmatrix}= \lambda\begin{bmatrix}a & b \\ c & d \end{bmatrix}= \begin{bmatrix}\lambda a & \lambda b \\ \lambda c & \lambda d\end{bmatrix}$.

So we must have $2c= \lambda a$, $2a+ 2d= \lambda b$, $\lambda c= 0$, and $\lambda d= 2c$. Look at $\lambda c= 0$. Either $\lambda= 0$ or $c= 0$.

If $\lambda$ is not 0 Then we must have $c= 0$ and the other equations become $\lambda a= 0$, $\lambda d= 0$, and $2a+ 2b= \lambda b$. With $\lambda$ not 0, that gives a= d= 0 and then b= 0. That is the "trivial solution" so that any non-zero number is not an eigenvalue. If $\lambda= 0$ then 2c= 0, and 2a+ 2d= 0, so that d= -a while b can be any number.

The only eigenvalue is 0 with corresponding eigenvectors of the form $\begin{bmatrix}a & b \\ 0 & -a\end{bmatrix}= a\begin{bmatrix}1 & 0 \\ 0 & -1\end{bmatrix}+ b\begin{bmatrix}0 & 1 \\ 0 & 0\end{bmatrix}$

Answer 3

To find the eigenvalues, we set up $$ BX + XB = \lambda X. $$ Then the equation becomes $$ (B-\lambda I) X + XB = 0. $$ Let $M_{ B-\lambda I}$ be the $\mathbb{C}[x]$-module provided by $B-\lambda I$ and $M_{-B}$ be the $\mathbb{C}[x]$-module provided by $-B$. Then the solution to this equation can be regarded as $$ \mathrm{Hom}_{\mathbb{C}[x]} (M_{ B-\lambda I}, M_{-B}). $$ Since we have $$ M_{ B-\lambda I} \simeq \mathbb{C}[x]/(x+\lambda)^2, \ \ M_{-B} \simeq \mathbb{C}[x]/x^2, $$ Then nonzero solutions exist if and only if $\lambda = 0$.

Moreover, in case $\lambda =0$, we have the following isomorphism as $\mathbb{C}[x]$-modules: $$ \mathrm{Hom}_{\mathbb{C}[x]} (M_{ B-\lambda I}, M_{-B})\simeq \mathbb{C}[x]/x^2. $$ Thus, the only eigenvalue of $f$ is $0$ with corresponding $2$-dimensional eigenspace. This shows that $f$ is not diagonalizable.