Lower and upper bound

Question

Let $V, \hat{V} \in \mathbb{R}^{p \times s}$ be semi-orthogonal matrices where $s < p$. Then how to prove trace($\hat{V}\hat{V}^TVV^T/s$) is lower bounded by 0 and upper bounded by 1 and achieves maximum iff $\hat{V}\hat{V}^T=VV^T$?

Answer 1

If $A=\hat{V}\hat{V}^T$ and $B=VV^T$, taking a suitable unitary $U^*AU=D\text{diag}(\underset{s\, \text{times}}{1,\cdots,1}, 0,\cdots 0)$. It is known that any principal submatrix $S_B$ of $B\ge 0$ has trace less than that of $B$ so $0\le\text{trace}(U^*AUU^*BU)\le \text{trace}(DU^*BU)= \text{trace}(S_{U^*BU})\le s$

The equality case for $\text{trace}(DU^*BU)= s$ is true if and only if $U^*BU$ has its lower $(p-s)\times (p-s)$ submatrix zero and since $U^*BU\ge 0$ it is of the form $\begin{pmatrix}S&0\\0&0\end{pmatrix}$, with $S$ unitary having all eigenvalues ones . Thus $D=U^*BU$ and that completes the proof.