If $(U,\Sigma,V)$ is a singular value decomposition of $A$, do the first $\text{rank}A$ columns of $V$ and $U$ form orthonormal bases? Let


*

*$m,n\in\mathbb N$

*$A\in\mathbb R^{m\times n}$ and $|A|:=\sqrt{A^TA}$

*$r:=\operatorname{rank}A$

*$\sigma_1>\cdots>\sigma_r>\sigma_{r+1}=\cdots=\sigma_n=0$ denote the singular values of $A$
We say that $(U,\Sigma,V)$ is a singular value decomposition of $A$ if


*

*$U\in\mathbb R^{m\times n}$ is a partial isometry;

*$\Sigma=\operatorname{diag}(\sigma_1,\ldots,\sigma_n)\in\mathbb R^{n\times n}$;

*$V\in\mathbb R^{n\times n}$ is orthogonal


and $$A=U\Sigma V^T\tag1.$$

Can we show that
  
  
*
  
*The first $r$ columns $(e_1,\ldots,e_r)$ and $(f_1,\ldots,f_r)$ of $V$ and $U$ are orthonormal bases of $\mathcal R(|A|)$ and $\mathcal R(A)$, respectively?
  
*$\mathcal N(U)=\mathcal N(A)$ (noting that $\mathcal N(A)=\mathcal N(|A|)$)?
  

 A: You have $$\tag1 AA^*=U\Sigma^2 U^*.$$ Taking the trace in $(1)$, 
$$
\operatorname{Tr}(\Sigma^2)=\operatorname{Tr}(AA^*)=\operatorname{Tr}(U\Sigma^2U^*)=\operatorname{Tr}(\Sigma^2 U^*U). 
$$
Then
$$
0=\operatorname{Tr}(\Sigma^2\,(I-U^*U))=\operatorname{Tr}(\Sigma\,(I-U^*U)^2\,\Sigma)
$$
As the trace is faithful, we get $\Sigma(I-U^*U)^2\Sigma=0$, and so $(I-U^*U)\Sigma=0$. So $$\tag2
\Sigma=U^*U\Sigma=\Sigma\,U^*U.
$$
Now
$$
A^*A=V\Sigma U^*U\Sigma V^*=V\Sigma^2\,V^*. 
$$
Now
$$
\ker A=\ker A^*A=\ker V\Sigma^2 V^*=\ker \Sigma V^*.
$$
So, taking orthogonals,
$$
\operatorname{ran} A^*=\operatorname{ran}V\Sigma.
$$
This shows that the first $r$ columns of $V$ span the range of $A^*$ (which is the same as the range of $|A|$). Going back to $(1)$, 
$$
\ker A^*=\ker AA^*=\ker U\Sigma^2\,U^*=\ker \Sigma U^*,
$$
so 
$$
\operatorname{ran} A=\operatorname{ran} U\Sigma,
$$
so the first $r$ columns of $U$ span the range of $A$. 
It is not true in general that $\ker A=\ker U$. For instance take 
$$
A=\begin{bmatrix} 0&0\\1&0\end{bmatrix} \,\begin{bmatrix} 1&0\\0&0\end{bmatrix} \,\begin{bmatrix} 0&1\\1&0\end{bmatrix} =\begin{bmatrix} 0&0\\0&1\end{bmatrix}. 
$$
