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There is a N*K semi-orthogonal matrix V, where N<K, satisfying $\mathbf{V}\mathbf{V}^T = \mathbf{I}_{N\times N}$.

D is a K*K diagonal matrix with positive entries.

My question is what the eigenvalues of matrix $\mathbf{VDV}^{T}$?

PS:There already has some tips that $\mathbf{VDV}^{T}$ has the same non-zero eigenvalues as $\mathbf{DV}^{T}\mathbf{V}$ and $\mathbf{V}^{T}\mathbf{V}$ has the same non-zero eigenvalues as $\mathbf{V}\mathbf{V}^{T}$ (N eigenvalues equal to 1).

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Extend $V$ to a $K\times K$ orthogonal matrix $U$. Then $B=VDV^T$ is the leading principal $N\times N$ submatrix of $A=UDU^T$. So, you are essentially asking a question about the eigenvalues of the principal submatrix $B$ of a Hermitian matrix $A$. In general, you cannot infer the eigenvalues of $B$ from the eigenvalues of $A$, but the former are bounded by the latter. See Eigenvalues of the principal submatrix of a Hermitian matrix .

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