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The vector space of matrices that commute with A

  1. Jul 18, 2013 #1
    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?


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

  2. jcsd
  3. Jul 19, 2013 #2


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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.
  4. Jul 19, 2013 #3


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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].
  5. Jul 29, 2013 #4
    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?
  6. Jul 31, 2013 #5
    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.
  7. Jul 31, 2013 #6


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    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:

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