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 $\mathcal N(A)=\mathcal N(|A|)$)?