Kernel of commutator in linear algebra

Intro: Let $[A,B]=AB-BA\in\mathbb{M}_{2,2}$. We will write $[A;B]$ as follows, $K_B(A)=[A,B]$ where we assume that $B$ is maintained and $A\mapsto K_B(A)$ as a mapping from $\mathbb{M}_{2,2}\rightarrow \mathbb{M}_{2,2}$.

(a) Show that $K_B:\mathbb{M}_{2,2}\rightarrow \mathbb{M}_{2,2}$ is linear for any $B\in \mathbb{M}_{2,2}$.

Answer: I've shown that $K_B$ is linear for any $B\in \mathbb{M}_{2,2}$. It was just a definition in the textbook.

(b) Describe $Ker(K_B)$ when we say $B=\begin{pmatrix} a & 0\\ 0 & b \end{pmatrix}$ where $a$ and $b$ are real numbers.

Answer: I've also shown that $Ker(K_B)$ is empty (or does not exist) when $a\neq b$ and $Ker(K_B)=\{\binom{0}{1},\binom{1}{0}\}$ when $a=b$.

To my real question now.

(c) Show that $B\in Ker(K_B)$ for any $B\in\mathbb{M}_{2,2}$. Show that $K_B$ is not surjective.

Any suggestion to question (c)? Thanks in advance.

Solution 1:

''(c) Show that $B∈Ker(K_B)$ for any $B∈M_{2,2}$. Show that $K_B$ is not surjective.''

First, since $K_B(B) = [B,B] = B^2-B^2=0$, it follows that $B$ lies in the kernel of $K_B$.

Second, since the mapping $K_B$ is linear and has a nontrivial kernel (for $B\ne 0$ zero matrix), it follows that the mapping cannot be surjective by the dimension formula for linear mappings $T:V\rightarrow V$: $\dim V = \dim \ker(T) + \dim image(T)$.

Solution 2:

Let me give alternative proof that $K_B$ is not surjective. Let $A$ be any $2\times 2$-matrix, then we get by the cyclicity of the trace $$ tr(K_B(A))=tr(AB)-tr(BA)=0.$$ This means that (thanks to ancient mathematician for suggesting an example that works for all characteristics) the matrix $$C=\begin{pmatrix} 1 & 0\\ 0 &0 \end{pmatrix}$$ cannot be in the image as $$ tr(C)=1\neq 0.$$