Determining if $\operatorname{rank}(A)$ is always equivalent to $\operatorname{rank}(A^t*A)$

Question

$\def\rk{\operatorname{rank}}$

Hi,

So I know the theorem:

$$ \dim( \ker(A) ) = \dim ( \ker(A^{t}A) ) $$

But I am not sure why this implies that the above is always true. The reason why I am struggling with this concept is because A is an $n\times m$ matrix and $A$ transpose multiplied by $A$ gives an $m\times m$ matrix. Doesn't this mean:

$$ \rk(A) = n - \dim ( \ker(A)) $$

$$ \rk(A^{t}A) = m - \dim ( \ker(A^{t}A) ) $$

This would would imply that the two are only equal when $m = n$. Can someone explain to me where I am going wrong?

Thanks

Answer 1

$\def\rk{\operatorname{rank}}$You are mistaken in applying the rank nullity theorem for $A$. The rank is the dimension of the domain of the linear map minus the dimension of its kernel. Since $A$ represents a linear map $\Bbb R^m\to\Bbb R^n$ (remember the order of sizes $n,m$ is reversed here) the dimension of the domain is $m$ rather then $n$, and $$\rk(A)=m-\dim(\ker(A)).$$