TITLE: Column Space and SVD
QUESTION [14 upvotes]: 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 $U$)

REPLY [2 votes]: 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 $A$.