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  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?

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3 Answers 3

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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.

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

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

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