6

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!

user642796
  • 52,188
KelseyKabob
  • 77
  • 2
    Do you know what the definition of a symmetric matrix is? – EuYu May 10 '13 at 04:21
  • 1
    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. –  May 10 '13 at 04:25
  • @ KelseyKabob , if you do not know how to solve this exercise, then it is because you do not work. At least, I hope so... –  Nov 08 '15 at 00:24

3 Answers3

22

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?

Alex Wertheim
  • 20,278
  • 11
    +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). – Jyrki Lahtonen May 10 '13 at 04:46
10

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.

Brian M. Scott
  • 616,228
  • (+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. – achille hui May 10 '13 at 16:48
0

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

Colton Fitzjarrald
  • 55