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Math Help - Need help solving for this matrix

  1. #1
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    Need help solving for this matrix

    FIRST LET ME SAY: I would like to be able to do this problem for myself, so if you want to actually do the problem, please don't post the answer right away so I can try it myself first.


    The problem is: suppose the vector x =
    [1
    0
    2]

    and A is a (3x3) matrix such that (Atranspose)(A)(x) =
    [2
    2
    0]

    what is || Ax || ^2


    So far I'm pretty much completely lost. I was thinking I could set (Atranspose)(A) equal to C and then solve for Cx=[2,2,0] but to be honest I wasn't sure even how to do that. I know that if x had been a square matrix I could have multiplied both sides by the inverse of x but obviously x isn't a square matrix.

    Help please!
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  2. #2
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    Think about the way matrices act inside inner products. You know that

    \|Ax\|^{2}=\langle Ax,Ax\rangle.

    Does this give you any ideas?
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  3. #3
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    No sorry Correct me if I'm wrong, but isn't there a way to solve for A, or at least (Atranspose)A?

    How would I go about solving Bx = C if I know what x and C are but x isn't square?
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  4. #4
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    Even using the fact that A^{T}A is symmetric won't give you enough equations to solve for the elements of A^{T}A.

    How would I go about solving Bx = C if I know what x and C are but x isn't square?
    You can't even write the equation Bx = C, if both B and C are matrices. The dimensions don't agree. The LHS is a vector, the RHS is a matrix.

    Look at this:

    \langle x,A^{T}Ax\rangle=x^{T}A^{T}Ax=(Ax)^{T}Ax=\|Ax\|^{2  }. What does that tell you?
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  5. #5
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    I'm a little confused with your use of
    \langle x,A^{T}Ax\rangle

    What exactly does that represent? I thought at first you were writing (x, a^{T}Ax) but that doesn't make sense. Perhaps something I haven't learned yet?

    Looking at what else you have written though, if  || A x || ^2 = (Ax)^{T} Ax then I could multiply both sides of my original equation by x^{T} and get an answer of: [2]
    Am I on the right train of thought here? Also, I haven't ever seen a theorem that would help me prove that  ||Ax||^2 = (Ax)^{T} Ax Is there an easy way to prove this?
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  6. #6
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    It's the inner product. In this case, inner product = dot product. I like the angled brackets, because it enables me to distinguish from ordered pairs, or intervals. Parentheses are used for so many darn things!

    So, I'm using the notation \langle \mathbf{x},\mathbf{y}\rangle\equiv \mathbf{x}\cdot \mathbf{y}=\mathbf{x}^{T}\mathbf{y}, where the last multiplication is matrix multiplication.

    Make sense?
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  7. #7
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    Ahhh yes that makes much more sense. Ok I guess the only part I'm confused about now is :
    \|Ax\|^{2}=\langle Ax,Ax\rangle.

    I'm assuming this comes from some theorem I missed. Is there a way to prove it using steps that I would already know?
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  8. #8
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    Well, it's the definition of Euclidean length, really. You have, for a vector \mathbf{x}=(x_{1},x_{2},\dots,x_{n}), that

    \|\mathbf{x}\|=\sqrt{x_{1}^{2}+x_{2}^{2}+\dots+x_{  n}^{2}}=\sqrt{\mathbf{x}\cdot\mathbf{x}}.
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  9. #9
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    oooooh, Ok I think I got it now.

    So for this problem all I would have to do is show that ||Ax|| ^2 = (Ax)^{T} Ax and with that, I would get that ||Ax||^2 = [2,2,0] * x^{T} . Correct?


    edit: oh and that [2,2,0] matrix should be the (3x1) matrix I had before. I didn't know how to make it a 3x1 and not a 1x3
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  10. #10
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    Almost, but not quite. You need to do

    (Ax)^{T}Ax=(x^{T}A^{T})Ax=x^{T}(A^{T}Ax).

    Now do you see what to do?
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  11. #11
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    I belive so. I think that's what I meant though. If we multiply the original equation by x^{T} then we can simplify down to (Ax)^T (Ax).
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  12. #12
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    Hmm. What's your final answer?
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  13. #13
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    [2]
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  14. #14
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    I wouldn't write it as a vector, but yes, that's correct.
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  15. #15
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    How did you get that answer?
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