Column Space and SVD

Since $A = USV'$, then its column space must be the same as the column space of $US$, since $V$ is invertible. And $S$ is a diagonal matrix and only the first $r$ diagonal entries of $S$ are nonzero, so check that only the first $r$ columns of $U$ "survive" being multiplied by $S$.

in fact you can write as the sum of $r$ rank one matrices $$A = s_1u_1v_1^T + s_2u_2v_2^T + \cdots+s_ru_rv_r^T$$ where $u_1,u_2,\ldots, u_r$ and $v_1,v_2,\ldots, v_r$ make up the columns of the orthogonal matrices $U,V$ and $r$ is the rank of $A.$ to show that $u_1,\ldots, u_r$ spans the column space of $A,$ look at $Ax.$ $$Ax = s_1(v_1^T x) u_1 + s_2(v_2^Tx)u_2 + \cdots+s_r(v_r^Tx)u_r$$

you can look at the row space of $A$ by computing $y^TA$ in the same way.

Note that $Us_i = \sigma_i u_i$ because $S$ is diagonal where $\sigma_i$ is the $i$th singular value and $u_i$ is the $i$th column of $U$. So $A v_i = \sigma_i u_i$, $i=1,2,\dots,r$ spans the column space of $A$.