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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.

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3 Come on... – Najib Idrissi Feb 22 '14 at 16:29
@nik I'm studying right out of a school text book and I had a question. That's what this place is for, no? Sometimes the book isn't very clear and other people can explain it better. – free_mind Feb 22 '14 at 16:30
Yes, this place is for questions, but usually a modicum of research is expected from people who ask questions. You don't have to use your book and MSE as your only sources of information. – Najib Idrissi Feb 22 '14 at 16:32
In any case looking up "symmetric matrix" on google would have given you the same answers as here. – Najib Idrissi Feb 22 '14 at 16:32
I think it seems obvious to you what one should google because you're familiar with the subject. This a brand new subject to me and understanding what you need to research is paramount to being able to conduct effective research. – free_mind Feb 22 '14 at 16:37
up vote 6 down vote accepted

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"?

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I know what symmetric means, I just didn't understand it in these terms. In any case, your comment really drove the point home, "it equals it own transpose." That made it crystal clear. Thank you – free_mind Feb 22 '14 at 16:28

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.

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In other words, the matrix is symmetric with respect to the main diagonal. – Roland Feb 22 '14 at 16:26

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.

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

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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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