I was reading Gilbert Strang's book and he says that if $A=USV'$ be the SVD of A ( assume square for the moment) then the nullspace of A is given by the last $n-r$ columns of V and the column space by the first $r$ columns of $U$.
I understand the null space part. If \begin{align} A&=USV'\\ AV&=US\\ [Av_1 \;Av_2\cdots Av_r\cdots Av_n]&=[Us_1 \;Us_2\cdots Us_r\cdots Us_n] \end{align} The columns $Us_{r+1}\cdots Us_n$ are $0$ vectors and correspond to (applicatoin of A on) linearly independent vectors $v_{r+1}\cdots v_n$. Thus, vectors $v_{r+1}\cdots v_n$ form the null space of A.
How do I prove the statement on column space? I am not even able to start. This is not homework by the way. Just wanted to prove everything in Strang myself for fun.
I feel it has to be inclusion proof where I show that any vector in the column space of A is in the column space of $\hat{U}$ and vice versa. ($\hat{U}$ is first r columns of $U$)
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 $S$.