I was listening to a lecture by Gilbert Strang and he said that in a SVD, the columns of $U$ are in the column space of $A$...
$$ \begin{aligned} A &= U \Sigma V^T \\ AV &= U \Sigma \end{aligned} $$
so we can clearly see by this calculation that U is a combination of the columns of $A$, but then he said that $V$ was in the rowspace of $A$ which I was having a hard time justifying because he did not give an explanation for it. If I try the same thing I see
$$ U^TA = \Sigma V^T $$
does multiplying from the left give a combination of the rows? or is there another way to justify this?
You can transpose both sides of your equation $U^T A = \Sigma V^T$ to see that $$ A^T U = V \Sigma^T. $$ This shows that the columns of $V$ are in the column space of $A^T$.