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Consider a generalized Eigenvalue problem $Av = \lambda Bv$ where $A$ and $B$ are square matrices of the same dimension. It is known that $A$ is positive semidefinite, and that $B$ is diagonal with positive entries.

It is clear that the generalized eigenvalues will be nonnegative. What else can one say about the eigenvalues of the generalized problem in terms of the eigenvalues of $A$ and the diagonals of $B$? Equivalently, what else can one say about the eigenvalues of $B^{-1}A$?

It seems reasonable (skipping over zero eigenvalues) that

$$ \lambda_{min}(B^{-1}A) \geq \lambda_{min}(A)/B_{max} $$

but I am unable to see how one could rigorously show this, and it is perhaps a conservative bound. Equivalently again, what could one say about the eigenvalues of

$$ B^{-1/2}AB^{-1/2} $$


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When you say "positive semidefinite" are you including the condition that $A$ is self-adjoint? (Also, does it have real entries?) – Qiaochu Yuan May 30 '12 at 23:41
Yes I am. $A$ is symmetric, real, and positive semidefinite. – John May 30 '12 at 23:51

Let $A$ be a symmetric, real, and positive-semidefinite $n \times n$ matrix. Consider the quadratic form $v^T Av$ on $\mathbb{R}^n$. By restricting if necessary to the complement of the nullspace of $A$, we may assume WLOG that $A$ is positive-definite.

Lemma: The minimal eigenvalue of $A$ is the minimal value of $v^T A v$ on the unit sphere $||v|| = 1$.

This is a consequence of the (proof of the) spectral theorem. Now, let $D = B^{-1/2}$ be a diagonal matrix with positive real entries. What is the minimal value of $v^T DAD v = (Dv)^T A (Dv)$ on the unit sphere? Well, $||Dv||$ is at least the smallest diagonal entry $d_{min}$ of $D$, so it follows that the minimal value is at least $d_{min}^2$ times the minimal eigenvalue of $A$. The conclusion follows.

$D$ being diagonal is not essential to this argument; it just needs to be invertible. In general $d_{min}$ needs to be replaced by the inverse of the operator norm $||D^{-1}||^{-1}$ and we need to consider $v^T D^T AD v = (Dv)^T A (Dv)$.

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Thanks! Very helpful. – John May 31 '12 at 0:07

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