I need help with a proof for my liner algebra class.

If $A$ is a square matrix, then $AA^T$ and $A^TA$ are symmetric.

I have no idea where to start!

  • 2
    $\begingroup$ Do you know what the definition of a symmetric matrix is? $\endgroup$
    – EuYu
    Commented May 10, 2013 at 4:21
  • 1
    $\begingroup$ Have you tried looking at specific examples? For instance, choose a $2\times 2$, $3\times 3$, etc. matrix and look at those examples to help you prove the general case. $\endgroup$
    – user59083
    Commented May 10, 2013 at 4:25
  • $\begingroup$ @ KelseyKabob , if you do not know how to solve this exercise, then it is because you do not work. At least, I hope so... $\endgroup$
    – user91684
    Commented Nov 8, 2015 at 0:24

3 Answers 3


Hint: take the transposes of $AA^{T}$ and $A^{T}A$. Use what you know about the transpose to show $(AA^{T})^{T} = AA^{T}$, and $(A^{T}A)^{T} = A^{T}A$

In particular, if you are still struggling:

$(AA^{T})^{T} = (A^{T})^{T}A^{T} = AA^{T}$

Can you do the second?

  • 11
    $\begingroup$ +1: Just to make sure that OP sees the light: the rule in use here is that for all matrices $X,Y$ $$(XY)^T=Y^TX^T.$$ Plug in $X=A$, $Y=A^T$ (and the other way around). $\endgroup$ Commented May 10, 2013 at 4:46

AWertheim has already given the nice answer, but I think it worth pointing out that even if you don’t see that little trick, you can still attack the problem successfully by brute force. Say $A=[a_{ij}]$ and $A^T=[\hat a_{ij}]$. Then we know that $\hat a_{ij}=a_{ji}$: the entry in row $i$, column $j$ of $A^T$ is $a_{ji}$, the entry in row $j$, column $i$ of $A$.

The entry in row $i$, column $j$ of $AA^T$ is

$$\sum_ka_{ik}\hat a_{kj}=\sum_ka_{ik}a_{jk}=\sum_ka_{jk}a_{ik}=\sum_ka_{jk}\hat a_{ki}\;,$$

which is the entry in row $j$, column $i$ of $AA^T$. That is, if $AA^T=[b_{ij}]$, then we’ve just shown that $b_{ij}=b_{ji}$ for all $i$ and $j$, which is exactly what it means for $AA^T$ to be symmetric.

  • $\begingroup$ (+1) I like this answer more because IMHO, this is more appropriate for the questioner. For someone who don't immediately see $AA^{T}$ is symmetric by definition, there will be issue how to justify $(XY)^T = Y^TX^T$ in general. $\endgroup$ Commented May 10, 2013 at 16:48

Taking the transpose of $ AA^T $ and using the reverse order law of transposition, $ (AB)^T = B^TA^T $, would be the simplest proof.


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