Prove that \((AB)^T = B^TA^T\) and that \(A^TA\) is always a symmetric matrix.

Recall that the transpose of a matrix is the switching of rows and columns. This gives us

\[\begin{split} (AB)^{T}_{ij} &= (AB)_{ji} \\ &= \sum\limits^{n}_{k=1} a_{jk}b_{ij}\\ &= \sum\limits^{n}_{k=1} b_{ij}a_{jk}\\ &= (B^TA^T)_{ij} \end{split}\]

which shows us that \((AB)^T = B^T A^T\). This logic can be directly applied to the second part of the question:

\[(A^TA)^T = A^T(A^T)^T = A^T A\]

So \(A^T A\) is by definition symmetric.