5
$\begingroup$

Prove ${\rm tr}\ (AA^T)={\rm tr}\ (A^TA)$ for any Matrix $A$

I know that each are always well defined and I have proved that, but I am struggling to write up a solid proof to equate them. I know they're equal.

I tried to show that the main diagonal elements were the same but if I say that $A$ is $n\times m$ then $$(AA^T)_{ii} = (a_{11}^2+\dots +a_{1m}^2) + \dots + (a_{n1}^2+\dots +a_{nm}^2)$$ and $$(A^TA)_{ii} = (a_{11}^2+\dots +a_{n1}^2) + \dots + (a_{1m}^2+\dots +a_{nm}^2)$$

$\endgroup$
  • 3
    $\begingroup$ please write up the pieces of your proof and some one will make it solid... good luck! $\endgroup$ – user87543 Jan 30 '14 at 6:35
  • 2
    $\begingroup$ rearrange terms. Both are the same expression. $\endgroup$ – voldemort Jan 30 '14 at 7:03
  • $\begingroup$ @b00n, re-read your formula. $\endgroup$ – Martín-Blas Pérez Pinilla Jan 30 '14 at 8:52
  • $\begingroup$ Thank you @Martín-BlasPérezPinilla! How naive of mine $\endgroup$ – b00n heT Jan 30 '14 at 8:55
  • $\begingroup$ Well, was true! :-) $\endgroup$ – Martín-Blas Pérez Pinilla Jan 30 '14 at 8:57
4
$\begingroup$

By $(AA^T)_{ii}$ you seem to mean the $i$-th term on the diagonal of $AA^T$. But instead what you've writen is already $\mathrm{tr}(AA^T)$. Which is ok. Now, you just have to realize that both sums are the same, up to the order of the addends -which doesn't matter.

For instance: you have $a_{11}^2$ on both sums, haven't you?

Also $a_{1m}^2$ appears on both sums. Also $a_{n1}^2$...

Write a few more terms on both sums. Or write all terms in the particular case $2\times 3$ and you'll see it.

$\endgroup$
  • 1
    $\begingroup$ Thank you, I have working for hours on hours so I sort of derped out and over looked the obvious. $\endgroup$ – e2DAeyePi Jan 30 '14 at 7:10
  • $\begingroup$ @e2DAeyePi To make it a little more concrete, try writing each expression as a double sum in $\Sigma$ notation. If you're consistent with how you form the sums, you should get the same for each expression except that the order of summation is swapped. $\endgroup$ – G. H. Faust Jan 30 '14 at 7:18
11
$\begingroup$

To give you an idea of how to properly write these sort of proofs down, here's the proof.

For a matrix $X$, let $[X]_{ij}$ denote the $(i,j)$ entry of $X$. Let $A$ be $m\times n$ and $B$ be $n\times m$. Then \begin{align*} \mathrm{tr}\,(AB) &= \sum_{i=1}^n[AB]_{ii} \\ &= \sum_{i=1}^n\sum_{k=1}^m[A]_{ik}\cdot[B]_{ki} \\ &= \sum_{k=1}^m\sum_{i=1}^n[B]_{ki}\cdot[A]_{ik} \\ &= \sum_{k=1}^m[BA]_{kk} \\ &= \mathrm{tr}\,(BA) \end{align*} Your question is a special version of this result with $B=A^\top$.

$\endgroup$
0
$\begingroup$

All approaches to the question made so far are pretty good. I just want to add a small observation that will make the proof solid.

trace($AA^T$) or trace($A^TA$) is sum of squares of all elements of the matrix $A$ (which is the same as the sum of squares of all elements of the matrix $A^T$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.