Define for a fixed $A \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ the mapping:

$$L_A : \mathbb{M}^{2 \times 2}(\mathbb{R}) \to \mathbb{M}^{2 \times 2}(\mathbb{R}) : X \mapsto AX-XA. $$

Give two linearly independent $X_1,X_2 \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ such that $L_A (X_1) = L_A (X_2) = 0$.

Conclude that $\dim(\ker(L_A))\geq 2$.

This is the question I was given. However, I'm not quite sure how to solve it. I know that $X=\lambda \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ with $\lambda \in \mathbb{R}$ are solutions. Sadly I can't find two that are linearly independent. What am I missing here?

I know that I want to find the $X \in \mathbb{M}^{2 \times 2}(\mathbb{R})$ that satisfy $AX=XA$. But this is all I got.

Thanks in advance for any help.


If $A$ is not a multiple of the identity, $A$ and $I$ belong to the kernel. If $A$ is a multiple of the identity, everything is in the kernel.

Always look for easy answers first.


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