What does it mean for $AA^T$ to be symmetric? What does it mean for $AA^T$ to be symmetric?
A question in my book says to show that $AA^T$ is symmetric so I took a very simple matrix to try and understand this:
$A=\begin{bmatrix} 2 \\ 8
 \\ \end{bmatrix}$
$A^T=\begin{bmatrix} 2 & 8
 \\ \end{bmatrix}$
$AA^T=\begin{bmatrix} 4 & 16 \\ 16 & 64
 \\ \end{bmatrix}$
But I don't understand how this is symmetric.  
 A: A square matrix $B$ is called symmetric if $B^T = B$. In other words, the $(i,j)$-th entry is the same as the $(j,i)$-th entry which indeed is the case with the matrix that you obtain.
A: The matrix
$$
\begin{bmatrix} 4 & 16 \\ 16 & 64
 \\ \end{bmatrix}
$$
is symmetric because it equals its own transpose:
$$\begin{bmatrix} 4 & 16 \\ 16 & 64
 \\ \end{bmatrix}^T = \begin{bmatrix} 4 & 16 \\ 16 & 64
 \\ \end{bmatrix}$$
Isn't that the definition of "symmetric"?
A: A matrix $A$ is symmetric if $A^T=A$ and notice that this can happen only for square matrix. Moreover we can easily see that
$$(A^T)^T=A\qquad;\qquad(AB)^T=B^TA^T$$
Now in your case since we have
$$(AA^T)^T=(A^T)^T A^T=AA^T$$ 
hence the matrix $AA^T$ is symmetric.
A: $(AA^{T})^T=(A^T)^TA^T=AA^T$
So, $AA^T$is symmetric.
A: For vectors $A$ you can write $AA^T$ using the superoperator notation as follows:
$$
\rm{ vec} AA^T= A^*\otimes A,
$$
where $\otimes$ deontes the Kronecker product. 
If we now look at the matrix repesentation of the Transposition superoperator
$$
\hat T = \pmatrix{1 &0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1}
$$
we recognize the bit reversal operation (this holds for dimensions of $A$ equal to a power of $2$, as far as I remember). $\hat T$ applied on $X\otimes Y$ gives:
$$
\hat T (X\otimes Y) = Y \otimes X
$$
So the fact that $AA^T$ is symmetric, means that $\rm{ vec} AA^T$ is a eigenvector (with eigenvalue $+1$) of the transposition superoperator.
