Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
1  
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
add comment

1 Answer

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:

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.