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Subspace question

  1. Dec 4, 2007 #1
    Let A be an nxn matrix and let H= {B E Mnxn|AB=BA}. Determine if H is a subspace of Mnxn.

    This was a test question that I got incorrect. I didn't like the way my teacher proved this afterwards, they said it IS a subspace of Mnxn. Any help in explaining how it could be would be greatly appreciated.
  2. jcsd
  3. Dec 4, 2007 #2
    First, you need to show that H is contained within M (obvious; no work needed but for rigor, you should at least state that H is contained within M). I am assuming here that M, the space of nxn matrices, includes the operations + and * (addition and multiplication). You will need to show that H is closed under addition (i.e. any two matrices in H added together should result in a matrix that is also in H) as well as multiplication. I'll show you one step to give you an idea of what's needed.

    Suppose B1 and B2 are matrices in H. We wish to show that (B1+B2) is also a matrix in H. To do so, we must verify that A(B1+B2) = (B1+B2)A. Let's start with the left hand side:

    A(B1+B2) = A B1 + A B2 = B1 A + B2 A

    (here we can replace A B1 with B1 A because B1 is an element of H, and hence satisfies AB1 = B1A; similarly for A B2).


    B1 A + B2 A = (B1+B2) A.

    Hence, (B1+B2) satisfies the condition A(B1+B2) = (B1+B2)A. Therefore, (B1+B2) is also in H, so H is closed under addition. You finish the rest by showing H is closed under multiplication.
  4. Dec 4, 2007 #3


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    H is constructed from elements of M, so clearly this is not necessary at all.

    The only necessity to showing something is subspace is to show that it is non-empty, closed under addition and scalar multiplication.

    What's the easiest way to show something is non-empty? Well, is the identity matrix in there?
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