Trace calculation in singular value decomposition

Question

An $m \times n$ matrix $S,$ of rank $s,$ can be expressed in the singular value decomposition (SVD) as $S=V \Lambda U^{\prime},$ where $\Lambda=\operatorname{diag}\left(\lambda_{1}, \ldots, \lambda_{s}\right)$ with $\lambda_{1}^{2} \geq$ $\cdots \geq \lambda_{s}^{2}>0$ being the nonzero elgenvalues of $S S^{\prime}, V=\left[V_{1}, \ldots, V_{s}\right]$ is an $m \times s$ matrix such that $V^{\prime} V=I_{s},$ and $U=\left[U_{1}, \ldots, U_{s}\right]$ is $n \times s$ such that $U^{\prime} U=I_{s} .$ Let $P=MN$ of rank $r(< s)$ where $M$ is an $m\times r$ matrix and $N$ is an $r\times n$ matrix. Then how to obtain the following equation: $$\operatorname{trace}\left[\left(V \Lambda U^{\prime}-P\right)\left(V \Lambda U^{\prime}-P\right)^{\prime}\right]=\operatorname{trace}\left[\left(\Lambda-V^{\prime} P U\right)\left(\Lambda-V^{\prime} P U\right)^{\prime}\right]$$ I have tried using $\operatorname{trace}(AB)=\operatorname{trace}(BA)$ but failed in the last term, namely $\operatorname{trace}(PP')=\operatorname{trace}(V^{\prime} P UU'P'V)$.

Answer 1

The claim is not true.

Take $$S=\begin{pmatrix}0&1&0\\0& 0& 1\end{pmatrix},\qquad P=\begin{pmatrix}0& 0& 1\\1& 0& 0\end{pmatrix}$$

Then $$S=VDU'=\begin{pmatrix}0&1\\1& 0\end{pmatrix}\begin{pmatrix}1& 0\\0&1\end{pmatrix}\begin{pmatrix}0&0&1\\0&1&0\end{pmatrix}$$

$$\mathrm{tr}(S-P)(S-P)'=\mathrm{tr}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}=4$$ $$\mathrm{tr}(D-V'PU)(D-V'PU)'=\mathrm{tr}\begin{pmatrix}1&0\\0&1\end{pmatrix}=2$$