The vector space of matrices that commute with A

  • Thread starter Bipolarity
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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
 

hilbert2

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Let's demonstrate this with 2X2 matrices...

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

[itex]a + 2c - a - 3b = 0[/itex]
[itex]b + 2d - 2a - 4b = 0[/itex]
[itex]3a + 4c - c - 3d = 0[/itex]
[itex]3b + 4d - 2c - 4d = 0[/itex]

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

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

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