$A$ is any real matrix. Is true that $\operatorname{Im}{(A^tA)}=\operatorname{Im}{(A^t)}$?
I only proved that $\operatorname{Im}{(A^tA)}\subset\operatorname{Im}{(A^t)}$.
$A$ is any real matrix. Is true that $\operatorname{Im}{(A^tA)}=\operatorname{Im}{(A^t)}$?
I only proved that $\operatorname{Im}{(A^tA)}\subset\operatorname{Im}{(A^t)}$.
You've shown that $\rm{Im}(A^\mathrm{T}A)$ is a subspace of $\rm{Im}(A^\mathrm{T})$. The general result follows by showing that $\rm{rank}(A) = \rm{rank}(A^\mathrm{T}A)$. This answer may be helpful to you.
What can you conclude about a subspace in a vector space of equal dimension?