Eigen-decompostion involved with a semi-orthogonal matrix

Question

Let $V$ be a $m$ by $n$ ($m>n$) semi-orthogonal matrix, i.e. $V^TV=I_n$
Can we say something about eigen-decomposition of $V^TPV$ where $P$ is a semi-positive definite matrix?
Result would be function of $V$ and eigen-values and eigen-vectors of $P$.
If not, can we say something about the smallest $d$ eigen-values of $V^TPV$ and the corresponding eigen-vectors provided that $d<n$?
I only think of utilizing Cauchy's interlacing property.
Let $U$ be a basis of orthogonal complement of span{$V$'s columns}, i.e. $[V ,U]$ be the basis of $\mathbb{R^m}$
$V^TPV$ is the principal $n$ by $n$ submatrix of $[V ,U]^TP[V ,U]$ and then apply the interlacing property. But the bound it provided is too loose and it doesn't tell anything about eigenvectors.
Any help would be appreciated!

Answer 1

As Exodd's comment mentions, Cauchy's interlacing property gives the best possible bounds. This is rigorously stated in Horn and Johnson's Matrix Analysis (second edition) as follows.

Theorem 4.3.21: Let $\lambda_1,\dots,\lambda_n$ and $\mu_1,\dots,\mu_{n+1}$ be real numbers that satisfy the interlacing inequalities $$ \mu_1 \leq \lambda_1 \leq \mu_2 \leq \lambda_2 \leq \cdots \leq \lambda_{n-1} \leq \mu_{n} \leq \lambda_{n} \leq \mu_{n+1}. $$ Let $\Lambda = \operatorname{diag}(\lambda_1,\dots, \lambda_n)$. A real number $a$ and a real vector $y \in \Bbb R^n$ may be chosen such that the eigenvalues of $$ A = \pmatrix{\Lambda & y\\ y^T & a} \in M_{n + 1}(\Bbb R) $$ are $\mu_1,\dots,\mu_{n+1}$.