Proof regarding the dimension of a vector space

Question

Fix a $2\times 2$ real matrix $A$. Let $V$ be the set of all $2\times 2$ real matrices $X$ such that $AX=XA$. Show that $V$ is a vector space of dimension of at least 2.

I'm struggling to see a good way to approach this problem. There's the brute force style method of algebraically manipulating 4 equations in 8 unknowns to show that there are (at least) two matrices $X$ that satisfy $AX=XA$ for any given $A$, but it seems like there should be a more insightful approach. Certainly the identity is in $V$, so there's one element in a basis. And the zero matrix is also in $V$, but this doesn't contribute to a basis as the columns are linearly dependent. And since we don't know that $A$ is invertible, we can't simply take $X=A^{-1}$.

Answer 1

Hint:

• Showing that it's a vector space should be easy enough without actually considering the entries of the matrices.

• To see that the dimension is at least $2$, consider two cases:

  • What if $A$ is a multiple of the identity matrix?
  • What if $A$ isn't a multiple of the identity matrix?

Answer 2

We can consider the following cases

• $A=0$ then X can be any matrix
• $A\neq 0$ singular then $X$ can be $kI$ and $kA$
• $A=kI$ then X can be any matrix
• $A\neq kI$ not singular then $X$ can be $kI$ and $kA$