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## Homework Statement

Prove ##tr(AA^T)=tr(A^TA)=s## where ##s## is the sum of the squares of the entries of A

I need help cleaning this up and I don't think my sigma notation is completely correct.

## The Attempt at a Solution

I found the identity $$(AB)^T=B^TA^T$$then applying it to ##AA^T \Rightarrow (AA^T)^T= (A^T)^TA^T=AA^T## and ##(A^T)^T=A, A^T=A^T## so now I know A is symmetric.

##A= \displaystyle\sum_{i,j=1}^{n} a_{i,j}## and ##A^T= \displaystyle\sum_{j,i=1}^{n} a_{j,i}##, ##A=A^T##so we know ##\displaystyle\sum_{j,i=1}^{n} a_{j,i}= \displaystyle\sum_{i,j=1}^{n} a_{i,j}##

also ##s=\displaystyle\sum_{i,j}^{n}(a_{i,j})^2##

then ##AA^T=A^TA=A^2 \Rightarrow [AA^T]_{i,k} = \displaystyle\sum_{j}^{n} a_{i,j}a_{j,k} ##

so ##Tr(A^2) = Tr\Bigg(\displaystyle\sum_{j}^{n} a_{i,j}a_{j,k}\Bigg) = \displaystyle\sum_{i,j}^{n}(a_{i,j})^2 = s ##

thanks!

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