# The vector space of matrices that commute with A

#### Bipolarity

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

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#### 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 $n^2$.

#### MisterX

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?

#### HasuChObe

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
Gold Member
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.
This almost cuts out the full heart of the matter. The matrices will commute if and only if there is a basis where they are both upper triangular.

There's even a wikipedia article about it:

http://en.wikipedia.org/wiki/Commuting_matrices

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