Given a square matrix $A$, both $AA^T$ and $A^TA$ are symmetric

Solution 1:

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?

Solution 2:

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.