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Thread: Symmetric matrix

  1. #1
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    Symmetric matrix

    If $\displaystyle A$ is a square matrix and $\displaystyle x^TAy=(Ax)^Ty$ for all $\displaystyle x,y\in R^n$, then prove $\displaystyle A$ is a symmetric matrix. ie. prove $\displaystyle A^T=A$

    I cant just go $\displaystyle x^TAy=x^TA^Ty$ rite? So what should i do instead?
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  2. #2
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    Use the matrix property $\displaystyle (AB)^T=B^TA^T.$
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  3. #3
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    yeh i know that property but how do i do it? ive just used the property and now im at $\displaystyle x^TAy=x^TA^Ty$. But i cant just cancel them out to get $\displaystyle A=A^T$ because they are matrices...
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  4. #4
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    You're told that the relationship is true for all $\displaystyle x$ and $\displaystyle y$, so it's certainly true for specific $\displaystyle x$ and $\displaystyle y.$
    Choose $\displaystyle x$ to consist of entirely of zeros apart from a 1 in the ith position and $\displaystyle y$ to consist entirely of zeros apart from 1 in the jth position. Then evaluate both sides of the equation $\displaystyle x^TA^Ty=x^TAy.$
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  5. #5
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    You started with:
    A is nxn
    x^TAy= (Ax)^Ty
    so
    x^TAy= x^TA^Ty
    x^TAy - x^TA^Ty = 0
    factor...
    x^T(A - A^T)y = 0
    this is true for all x,y in R^n
    consider x = 0, and y = anything in R^n we get x^T(A - A^T)y = 0
    consider y = 0, and x = anything in R^n we get x^T(A - A^T)y = 0
    consider y=x=0 we get x^T(A - A^T)y = 0
    we know the statement is true for all x,y not just x, y, being zero vectors
    the only way x^T(A - A^T)y = 0 is true for any x and y is if A- A^T = zero matrix

    hope this helped
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