What does it mean for $AA^T$ to be symmetric?

Question

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.

Answer 1

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"?

Answer 2

$(AA^{T})^T=(A^T)^TA^T=AA^T$ So, $AA^T$is symmetric.

Answer 3

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.

Answer 4

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.