What does it mean for $AA^T$ to be symmetric?
Andrew Henderson 
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.
$\endgroup$ 65 Answers
$\begingroup$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"?
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ 1 $\begingroup$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.
$\endgroup$ $\begingroup$$(AA^{T})^T=(A^T)^TA^T=AA^T$ So, $AA^T$is symmetric.
$\endgroup$ $\begingroup$For vectors $A$ you can write $AA^T$ using the superoperator notation as follows: $$ \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.
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