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I am trying to prove this problem and I am having trouble were to exactly start. The question is:

Show that if $A$ is any matrix, then $K = A^T A$ and $L = AA^T$ are both symmetric matrices.

My attempt:

In order to be symmetric then $A=A^T$ then $K = AA$ and since by definition we have that $K = A^n$ is symmetric since $n > 0$.

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If you have matrices $B,C,$ what are $(BC)^T$ and $(B^T)^T?$ – Will Jagy Sep 21 '12 at 0:46
You confuse the variable $A$ in the definition of symmetry with your matrix $A$. Don't do it :D – Long Sep 21 '12 at 0:50
up vote 6 down vote accepted

You need to use $(XY)^T=Y^TX^T$. Apply this identity and you'll get it.

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Okay, I think I got it. Thank you for your input! – diimension Sep 22 '12 at 2:08
actually I did not get. I might understand it better if you can tell how to start it off? – diimension Sep 24 '12 at 6:45
@diimension Let $X=A$ and $Y=A^T$ note that $Y^T=(A^T)^T=A$ hence $(AA^T)^T=(A^T)^TA^T=AA^T$ hence $L=AA^T$ is a symmetric matrix. The proof for $A^TA$ is much the same. The bigger point here is that you need to ask: "what do I need to show?" You need to show $L^T=L$ and $K^T=K$. When you begin by putting $A^T=A$ you assume something about $A$ which was not given... Long was warning you of this pitfall... hope this helps. – James S. Cook Sep 25 '12 at 0:50
Yup, thank you very much James! – diimension Sep 25 '12 at 6:13

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