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Suppose we have a full rank positive definite Hermitian matrix $A\in \mathbb{C}^{n \times n}$ with eigenvalues $\lambda_1>\lambda_2> \dots >\lambda_n$. Consider a semi-orthogonal matrix $X \in \mathbb{C}^{n \times p}$ (i.e., $X^* X = I_p$) spanning a subspace of $A$. Denote the eigenvalues of matrix $B = X^* A X$ to be $s_1 > s_2 > \dots > s_p$.

Is it possible to show that $\lambda_j \geq s_j, \forall 1\leq j \leq p$?

I ran through a simulation for $10^5$ trials over $n=10$ and $p=4$, and the results suggested so. But I cannot prove it or raise a counter example.

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What do you mean when you write about a matrix spanning a subspace? What do you mean by a subspace of a matrix? – Gerry Myerson Jul 22 '13 at 9:14
@GerryMyerson Perhaps a confusion here. I meant X would be used to project A onto a lower dimensional subspace such that B=X*AX has dimension p. – Allen Chen Jul 22 '13 at 12:39
Confusion persists. You continue to attribute to matrices properties belonging to vector spaces. What do you mean by the dimension of a matrix? – Gerry Myerson Jul 22 '13 at 12:45
up vote 0 down vote accepted

Let $X = U {\rm I}_{n \times p} V^*$ be an SVD of $X$: $U$ is unitary of order $n$, $V$ is unitary of order $p$, and ${\rm I}_{n \times p}$ is diagonal of order $n \times p$ with all diagonal elements equal to $1$. Then

$$B = X^* A X = V {\rm I}_{p \times n} U^* A U {\rm I}_{n \times p} V^*,$$

so $B$ is (unitarily) similar to $B' := {\rm I}_{p \times n} U^* A U {\rm I}_{n \times p}$, which means that $B$ and $B'$ have the same eigenvalues. However, $A' := U^* A U$ is (unitarily) similar to $A$, so they have the same eigenvalues as well.

Notice that ${\rm I}_{p \times n} A' {\rm I}_{n \times p}$ is the top left principal submatrix of $A'$, so you can apply Cauchy interlacing theorem, which will give you a bit more than you asked for:

$$\lambda_j \ge s_j \ge \lambda_{n-p+j}, \quad \text{for $1 \le j \le p$}.$$

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