# The vector space of matrices that commute with A

Suppose ##A## is a ## n \times n## matrix.

Define the set ## V = \{ B | AB = BA, B \in M_{n \times n}( \mathbb{F}) \} ##
I know that ##V## is a subspace of ##M_{n \times n}( \mathbb{F}) ## but how might I go about finding the dimension of ##V##? Is this even possible? It seems like an interesting problem, but constructing a basis for ##V## seems to me challenging enough. Any tips for me?

Thanks!

P.S. Not a homework problem, I made it myself and not sure if it has a simple answer.

BiP

Related Linear and Abstract Algebra News on Phys.org
hilbert2
Gold Member
Let's demonstrate this with 2X2 matrices...

Let A = $\left( \begin{array}{cc} 1 & 2 \\ 3 & 4 \\ \end{array} \right)$ . Multiplying this from either side with the general 2X2 matrix $\left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right)$ and requiring that the matrix elements of these products are equal, we get a system of linear equations:

$a + 2c - a - 3b = 0$
$b + 2d - 2a - 4b = 0$
$3a + 4c - c - 3d = 0$
$3b + 4d - 2c - 4d = 0$

Now we get the dimensionality of the space of matrices commuting with A as the dimensionality of the kernel of the coefficient matrix of this linear system.

HallsofIvy
Homework Helper
The answer is going to depend upon A. For example, if A= 0, AB= BA= 0 for all B so the space of all matrices that commute with A has dimension $n^2$.

In general you would have $n^2$ equations
$$\sum_{m=1}^n A_{im} B_{mj} - A_{mj}B_{im} = 0$$
The dimension of $V$ will be a most $n^2$ and at least $2$ (providing $n > 1$), since all $n\times n$ matrices commute with scalar multiples of the identity, as well as scalar multiples of themselves. Perhaps it is interesting to consider additional constraints on $\mathbf{A}$ and $\mathbf{B}$. For example, what if $\mathbf{A}$ and $\mathbf{B}$ are Hermitian?

I'm not a very mathy person, so I'll just drop my 2 cents. If B has the same eigenvectors as A, then it commutes with A.

Office_Shredder
Staff Emeritus