How can I prove that $Tr(A^TA) = \sum_{i=1}^n\sum_{j=1}^m a_{ij}^2$ using sigma sum notation?

Question

See above^. Original question is to show that the Trace is equal to the sum of the squares of all the elements

I particularly do not understand, in the following proof, the the steps 2 and 3 and how we know that $\sum_{j=1}^{n}(A^TA)_{jj} = \sum_{i=1}^n\sum_{j=1}^m a_{ji}^Ta_{ij}$ and how we know that $\sum_{i=1}^n\sum_{j=1}^m a_{ji}^Ta_{ij} = \sum_{i=1}^n\sum_{j=1}^m a_{ij}^2$. I'd be grateful if someone could explain,

$Tr(A^TA) = \sum_{j=1}^{n}(A^TA)_{jj}$

$\therefore \sum_{j=1}^{n}(A^TA)_{jj} = \sum_{i=1}^n\sum_{j=1}^m a_{ji}^Ta_{ij}$

$\therefore =\sum_{i=1}^n\sum_{j=1}^m a_{ij}^2$

Answer 1

If $\;A=(a_{ij})\;,\;\;1\le i\le n\;,\;\;1\le j\le m\;$ , then $\;A^t=(b_{ij})\;$ ,with $\;b_{ij}=(a_{ji})\;$ , so

$$AA^t=\left(\sum_{k=1}^ma_{ik}b_{kj}\right)=\left(\sum_{k=1}^m a_{ik}a_{jk}\right)\implies \text{Tr.}\,\left(AA^t\right):=\sum_{i=1}^m\sum_{k=1}^na_{ik}a_{ik}=\sum_{i=1}^m\sum_{k=1}^na_{ik}^2$$