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Math Help - Help with some matrix proofs

  1. #1
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    Help with some matrix proofs

    I am having a hard time with proving some statements that deal with matrix operations.


    Here's what I'm working on.

    1) A and B are both n by n matrices.

    Show that the formula (A + B)^2 = A^2 + 2AB + B^2 is not valid in general....

    Here's what I worked up so far....

    On the LHS you have AA + AB + BA + BB when you foil out (A + B)^2

    On the RHS you have AA + AB +AB + BB when you express the squares without exponents.

    It simply isn't good enough to check a single matrix and show that

    AA + AB + BA + BB is not equal to AA + AB + AB + BB

    but I do know that AB and BA will have different results.



    Secondly to prove the distributive property of matrices :

    (c + d)A = cA +dA

    Let A be an m x n matrix and let c and d be scalars

    not really sure what to do here at all....


    Lastly to prove the statement (A + B)C = AC + BC

    Let A = m x n matrix, B = m x n matrix and C = n x p matrix


    I'm even more unsure of how to prove this.... I would think that

    A and B are of the same dimensions such that for example A and B are 3x2
    matrices and C would be a 2x4 matrix... your resulting matrix would be 3x4 ?

    Even still that has nothing to do with the actual proof ?


    Anyways I would REALLY appreciate any help!
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  2. #2
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    Quote Originally Posted by battleman13 View Post
    1) A and B are both n by n matrices.

    Show that the formula (A + B)^2 = A^2 + 2AB + B^2 is not valid in general....

    Here's what I worked up so far....

    On the LHS you have AA + AB + BA + BB when you foil out (A + B)^2
    Can't you just end it here?

    Given matrix multiplication 'in general' does not commute, then say as

    \displaystyle AB \neq BA \implies (A + B)^2 \neq A^2 + 2AB + B^2
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  3. #3
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    Quote Originally Posted by battleman13 View Post
    Lastly to prove the statement (A + B)C = AC + BC

    Let A = m x n matrix, B = m x n matrix and C = n x p matrix


    I'm even more unsure of how to prove this.... I would think that

    \displaystyle (A_{mxn}+B_{mxn})C_{nxp}=A_{mxn}C_{nxp}+B_{mxn}C_{  nxp}=(AC)_{mxp}+(BC)_{mxp}=AC+BC
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  4. #4
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    Quote Originally Posted by pickslides View Post
    Can't you just end it here?

    Given matrix multiplication 'in general' does not commute, then say as

    \displaystyle AB \neq BA \implies (A + B)^2 \neq A^2 + 2AB + B^2
    I think I perhaps can but I would need to "show" that matrix multiplication does not commute not for one case but every case.

    At any rate thank you both very much for your responses!
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  5. #5
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    Quote Originally Posted by battleman13 View Post
    I think I perhaps can but I would need to "show" that matrix multiplication does not commute not for one case but every case.

    At any rate thank you both very much for your responses!
    Just stating matrix multiplication doesn't commute and then saying therefore AB + BA doesn't equal 2AB will be fine. It only commutes for special cases.

    Since you have the word 'in general', people will assume this isn't a special case.
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  6. #6
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    Proving that an identity isn't valid in general means exactly that; there may be specific cases for which it is valid. You do only need to find one matrix for which the identity doesn't hold. You can't prove that matrix multiplication doesn't commute in every case because it's not true. I and -I, where I is the identity matrix of any order, commute.

    As for associativity, there is another way. You can show that matrix multiplication corresponds (pretty much by definition) to composition of linear transformations, which are just special functions, and that function composition is associative. It's a more elegant way that avoids messy computations but it still takes some work.
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  7. #7
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    Quote Originally Posted by battleman13 View Post
    I think I perhaps can but I would need to "show" that matrix multiplication does not commute not for one case but every case.
    You can't do that because it is not true- there do exist matrices A, B such that AB= BA.

    At any rate thank you both very much for your responses!
    In order to prove that a general statement is not true, you only have to show a single "counter example".

    Since, "commuting matrices" are, at any rate, rare, you might try just picking matrices at random. For example, suppose
    A= \begin{bmatrix}2 & 1 \\ 1 & 3\end{bmatrix}
    and
    B= \begin{bmatrix}3 & 2 \\ 2 & 4\end{bmatrix}
    (which I did, literally, choose at random)

    What is A+ B? What is (A+ B)^2? Is it the same as A^2+ 2AB+ B^2? If not, you are done.
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