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Matrix manipulation/arithmetic in equations for proofs

  1. Apr 14, 2013 #1
    Hi there again guys!

    I didnt really know what to call this thread, because my problem isnt actually to do with how to manipulate the elements of the matrix itself, but rather how to deal with the actual symbol for the matrix in equations.

    I'll start off with a fundamental thing, even though I'm not sure it makes sense. Suppose I have a matrix equation AB-1= C. Obviously the order matters unlike in regular algebra, so if I rearrange this for A, do i get A = BC or A = CB? how do I tell the order it needs to be in, generally?

    Now I'll go onto a specific example. I need to prove that (AT)-1 = (A-1)T. My text book makes sense until the last step. Here's what it says if I fill in some gaps:

    1) Start with AA-1 = I = A-1A, where I is the identity matrix.
    2) Transpose each term. So (AA-1)T = IT = (A-1A)T
    3) Take AT out of the brackets: (A-1)TAT = AT(A-1)T.
    4) Then it says "Clearly (A-1)T = (AT)-1"... I dont really get how that comes about, feels like it got plucked out of nowhere, unless im missing something stupid. I think mainly my problem is that im pretty new to this whole thing so I dont understand some basic operations. So if you guys know of some material online that maybe goes over this sort of thing, I'd love to check it out.

    Thanks guys!
     
  2. jcsd
  3. Apr 14, 2013 #2
    AB^-1 = C

    multiply each side by B from the right,

    (AB^-1)B = CB

    Use the identity B^-1*B = I to say A = CB.

    It's important to make sure that when you multiply each side of an equation by a matrix, you multiply it from the correct direction, eg. if A = B, AC =/= CB in general. If you multiply both sides by C, the C must either go on the far left end or on the far right end. CA = CB and AC = BC. In short, matrix multiplication is non-commutative, so the order of operations matters.

    Regarding the proof... in step 3, they lose the "= I =" middle section, but both LHS and RHS still equal the identitiy matrix, so what you really have is two separate expressions that both equal I (I^T = I). If you write these down and compare them to the definition of the matrix inverse, it should be exactly enough proof to show that (A^T)^-1 = (A^-1)^T
     
  4. Apr 14, 2013 #3

    mfb

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    Remember how you get rid of B-1: You multiply both sides with B on the right side and get:
    AB-1B= CB
    Afterwards, you can use B-1B = I and get A=CB
    If you would multiply with B on the left side, you would get BAB-1= BC, but that cannot be simplified in the same way.
    No need to guess the result, you can derive it step by step.

    In steps 3-4, remember that both sides are still equal to I. Therefore, you found a matrix which, multiplied by AT, gives I - it is the inverse of AT.
     
  5. Apr 14, 2013 #4
    These two posts are almost word for word and at the same time. It's nice that someone verifies I'm not talking rubbish every now and then!
     
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