Why is the following true? $\operatorname{rank}(A^G A) = \operatorname{rank}(A)$

Question

I saw the following theorem in my text book:

If $A$ is an $n \times m$ matrix with $n \geq m$ then

$$\operatorname{rank}(A^G A) = \operatorname{rank}(A)$$

Note: $A^G$ here is the generalized inverse of $A$.

I'm trying to understand why this is and would really appreciate it if someone could prove it.

Answer 1

Because by definition of $A^G$, we have $AA^GA=A$ and by the general properties of rank with respect to products, the rank of $A^GA$ lies between the rank of $A$ and the rank of $AA^GA$.