# 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?

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 symmetric 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.

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

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