Please tell me a simple proof for the following equality: $$\operatorname{rank} A=\operatorname{rank} A A^T.$$
I proved the above equality as follows:
I found the following formula in "Linear Algebra" (in Japanese) by Ichiro Satake.
Let $V, V^{'}$ be finite dimensional vector spaces.
Let $f:V\to V^{'}$ be a linear mapping.
Let $W\subset V$ be a subspace.
Then, $\dim f(W) = \dim W - \dim(f^{-1}(0)\cap W)$.
I know $N(A)\cap C(A^T)=\{0\}$ from Gilbert Strang's book, where $N(A)$ is the nullspace of $m\times n$ matrix $A$ and $C(A^T)$ is the row space of $m\times n$ matrix $A$.
I combined the above two propositions as follows:
$$\dim L_A(C(A^T))=\dim C(A^T)-\dim (N(A)\cap C(A^T)).$$
Since $\dim L_A(C(A^T))=\dim C(A A^T)$ and $\dim (N(A)\cap C(A^T))=0$, $$\dim C(A A^T)=\dim C(A^T).$$
This means $$\operatorname{rank} A=\operatorname{rank} A A^T.$$
Since $\operatorname{rank} A = \operatorname{rank} A^T$, $$\operatorname{rank} A=\operatorname{rank} A A^T=\operatorname{rank} A^T A.$$
