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

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?

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

$$\newcommand{\rank}{\operatorname{rank}}$$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 $$\rank(AA^T)=\rank(A)$$.

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.

• Or, in other words, relabel $A$ as, say, $M$ in the formulation of the given theorem, and then use $M=A^T$.
– user491874
Dec 2, 2017 at 17:03
• thank you. what about the 2nd problem? Dec 3, 2017 at 1:52
• 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. Dec 3, 2017 at 2:03

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 $$\mathrm{rank}(AA^T) = 0 < 1 = \mathrm{rank}(A)$$.