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Proving AA^T is Symmetrical

  1. May 7, 2014 #1
    1. The problem statement, all variables and given/known data

    Mary Boas 3.9.5

    Show that the product AA^T is a symmetric matrix.

    2. Relevant equations

    $${ \left( AB \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } { B }_{ kj }$$

    3. The attempt at a solution

    I do not have a solution for this one, so could someone please check my work? I never took geometry or was required to do any proofs in calculus, so go easy on me.

    If $${ \left( AA \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } A_{ kj }$$, then $${ \left( A{ A }^{ T } \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ ik } } A_{ jk }$$.

    ∴ $${ \left( A{ A }^{ T } \right) }_{ ij }=\sum _{ k }^{ }{ { A }_{ jk } } A_{ ik }=\sum _{ k }^{ }{ { A }_{ jk } } { A }_{ ki }^{ T }={ \left( A{ A }^{ T } \right) }_{ ji }$$

    Since, $$ { \left( A{ A }^{ T } \right) }_{ ij }={ \left( A{ A }^{ T } \right) }_{ ji }$$, $$A{ A }^{ T } $$ is a symmetric matrix.
     
  2. jcsd
  3. May 7, 2014 #2

    jbunniii

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    Yes, it looks fine to me. Using exactly the same method of proof, you can show that ##(XY)^T = Y^T X^T## for any matrices ##X## and ##Y## of the appropriate size for the product to make sense. The symmetry of ##AA^T## is a special case of this result: set ##X = A## and ##Y = A^T##.
     
  4. May 7, 2014 #3
    You cannot have ##A_{ki}^T## because ##A_{ki}## is a term in the matrix. Note however that the terms commute so

    ##A_{jk}A_{ik}=A_{ik}A_{jk}##

    ##(AA^T)_{ij}=\sum_kA_{jk}A_{ik}=\sum_kA_{ik}A_{jk}=(AA^T)_{ji}##
     
  5. May 7, 2014 #4

    jbunniii

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    I assumed he meant ##(A^T)_{ki}##, the element in row ##k## and column ##i## of ##A^T##, which is of course the same as ##A_{ik}##.
     
  6. May 7, 2014 #5
    Okay, then its fine.
     
  7. May 7, 2014 #6
    Yes, indeed that is what I meant, but I did not know how to express it.

    Chris
     
  8. May 8, 2014 #7
    A friend of mine suggested $${ { (AA }^{ T }) }^{ T }={ A }^{ T }A$$ which is incorrect, but it led me to the MUCH more elegant proof using the "Relevant EQ" above to get $${ { (AA }^{ T }) }^{ T }={ AA }^{ T }$$. I think that does the job very nicely.

    Regards,
    Chris Maness
     
  9. May 8, 2014 #8

    D H

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    Both correct and nice. Very nicely indeed.
     
  10. May 8, 2014 #9
    Thank you for the encouragement. I have been working my butt off in this book. I started chapter 1 in the middle of February and I am only in Chapter 3, but I am doing every problem I have a solution and/or answer for. Chapter 3 is hard work because I have never had linear algebra.

    Chris
     
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