5
$\begingroup$
  1. Let $A$ be an $m\times n$ matrix. Prove that $\operatorname{rank}(AA^T) = \operatorname{rank}(A)$.

The problem tells me to prove it with the theorem that $\operatorname{rank}(A^TA) = \operatorname{rank}(A)$.

I'm a bit lost here...$AA^T$ and $(A)$ don't even have the same number of columns. I'm thinking maybe to prove it by showing that $[m - \operatorname{nullity}(AA^T)] = [n - \operatorname{nullity}(A)],$ but then I'm stuck here.

  1. Let A be an $m\times n$ matrix. Prove that the column space and row space of $A^TA$ are the same.

The problem tells me to prove it also with the theorem - $\operatorname{rank}(A^TA) = \operatorname{rank}(A)$. But I'm really running out of ideas.

Help?

$\endgroup$
2
  • $\begingroup$ Do you know the singular value decomposition? (SVD) $\endgroup$
    – Surb
    Dec 2, 2017 at 16:10
  • $\begingroup$ Proper notation is $n\times m,$ not $n$ x $m.$ I edited accordingly and also did some other copy-editing. $\endgroup$ Dec 2, 2017 at 16:33

3 Answers 3

3
$\begingroup$

$\newcommand{\rank}{\operatorname{rank}}$Considering $A$ as a linear map, $\operatorname{Im}(AA^T)\subset \operatorname{Im}(A)$ implies that $\rank(AA^T)\leq \rank(A)$.

Consider the canonical scalar product associated to the basis used to write the matrix. Then, $AA^T(x)=0$ implies that $\langle AA^T(x),x\rangle=\langle A^T(x),A^T(x)\rangle=0$ implies that $A^T(x)=0$. this implies that $\ker(AA^T)\subset \ker(A^T)$. since $\rank(A^T)=\rank(A)$ and $\rank(A)+\dim\ker(A)=n$, we deduce that $\rank(AA^T)=\rank(A)$.

$\endgroup$
2
$\begingroup$

Are you saying that:

  • There is a theorem that, for every $m\times n $ matrix $A$, $\operatorname {rank} (A^TA)=\operatorname {rank} (A) $,
  • You are supposed to use that theorem to prove that, for every $m\times n $ matrix $A$, $\operatorname {rank} (AA^T)=\operatorname {rank}(A) $

Assuming that is what it is:

Proof: Note the above theorem is valid for every matrix $A$. Pick a matrix $A $. Apply the above theorem to $A^T$. Thus, $\operatorname {rank}(A^{TT}A^T) = \operatorname {rank}(A^T) $. Then, use the facts that $A^{TT}=A $ and $\operatorname {rank}(A^T)=\operatorname {rank}(A) $ to reach the conclusion.

For the 2nd part, note the identity $(AB)^T=B^TA^T $, so $(A^TA)^T=A^TA^{TT}=A^TA $, i.e. $A^TA $ is a symmetrical matrix. Thus, not only the row and column spaces are the same - with the appropriate identification of rows to columns - but, with that identification, the rows and columns of $A^TA $ are the same.

$\endgroup$
3
  • $\begingroup$ Or, in other words, relabel $A $ as, say, $M $ in the formulation of the given theorem, and then use $M=A^T $. $\endgroup$
    – user491874
    Dec 2, 2017 at 17:03
  • $\begingroup$ thank you. what about the 2nd problem? $\endgroup$
    – Tommy Ling
    Dec 3, 2017 at 1:52
  • $\begingroup$ Let $A$ be an $m\times n$ matrix. Prove that the column space and the row space of $(A^T)A$ are the same. My thought is that since both column and row spaces of $(A^T)A$ are in $\Bbb{R}^n$, and the dimensions of both column and row spaces are the same, it follows that the column space and the row space of $(A^T)A$ are the same? But it somehow seems that I didn't successfully show that both column and row spaces $span$ $\Bbb{R}^n$. What do you think? Also, the problem tells me to use $rank((A^T)A)$ = $rank(A)$, a theorem that I did not use in the proof. $\endgroup$
    – Tommy Ling
    Dec 3, 2017 at 2:03
1
$\begingroup$

As a caveat, you must state that your matrix is over the field $\mathbb{R}$. If it is over $\mathbb{C}$, then the conclusion is not true. For example, take \begin{align*} A = \begin{pmatrix} 1 & i \end{pmatrix}, \end{align*} then $AA^T = 0$, whence $\mathrm{rank}(AA^T) = 0 < 1 = \mathrm{rank}(A)$.

$\endgroup$

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.