Find the Dimension of a Subspace of Matrices

[Euge]

Given a complex matrix $A\in M_n(\mathbb{C})$, let $X_A$ be the subspace of $M_n(\mathbb{C})$ consisting of all the complex matrices $M$ commuting with $A$ (i.e., $MA = AM$). Suppose $A$ has $n$ distinct eigenvalues. Find the dimension of $X_A$.

 

[Korybut]

Spoiler

Suppose that you found the basis where $A$ is diagonal and it looks as
$A=\operatorname{diag}\{\lambda_1,...,\lambda_n\}$

Then the commutativity condition in index notation turns into very simple one
$(\lambda_k-\lambda_i)M^i_k=0$
Here $i$ and $k$ run from $1$ to $n$. For $i=k$, $M^i_i$ can be any, however all other values must zero. Hence you have the space of diagonal matrices, which dimension is $n$.

 

[Office_Shredder]

Spoiler

Suppose that you found the basis where $A$ is diagonal and it looks as
$A=\operatorname{diag}\{\lambda_1,...,\lambda_n\}$

Then the commutativity condition in index notation turns into very simple one
$(\lambda_k-\lambda_i)M^i_k=0$
Here $i$ and $k$ run from $1$ to $n$. For $i=k$, $M^i_i$ can be any, however all other values must zero. Hence you have the space of diagonal matrices, which dimension is $n$.

I've always felt like it's a flaw of linear algebra that more matrices don't commute with diagonal matrices