Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Are there any techniques for lower-bounding the angle between eigenvectors of a matrix? Or a lower bound on the related quantity of the condition number of the matrix of eigenvectors? In particular I'm looking for bounds that depend on the difference in the corresponding eigenvalues, with larger angles when the eigenvalues are more separated.

For symmetric matrices (and more generally normal matrices) the angles are of course all right angles. I'm looking for techniques that apply to non-normal matrices.

(The particular class of matrices that I care about is stochastic matrices with trace $n-1$ as described in my previous question Bounds on $||P^{k+1} - P^k||$ for $n$ by $n$ stochastic matrix $P$ with trace $n-1$ and integer $k>>n$. .)

share|improve this question
    
Just a thought: for invertible matrices, I wonder if a relevant quantity is the difference between $||A^{-1}||_2$ and the spectral radius of $A^{-1}$. Consider: if $A$ is symmetric (so that eigenvectors are at right angles), then so is $A^{-1}$, and the spectral radius of $A^{-1}$ equals its norm. On the other hand, suppose we keep all the eigenvalues of $A$ the same but let the angle between two eigenvectors approach zero. Then the spectral radius of $A^{-1}$ does not change, but its norm approaches infinity. –  alex Nov 8 '10 at 22:56

1 Answer 1

Here's a generalization of the result that eigenvectors of a real symmetric matrix with distinct eigenvalues are orthogonal:

Let $x$ and $y$ be unit eigenvalues of a real matrix $A$ with eigenvalues $\lambda$ and $\mu$, respectively. Then $$ \lambda\langle x,y\rangle =\langle Ax,y\rangle =\langle x,A^\mathrm{T}y\rangle =\langle x,Ay\rangle-\langle x,(A-A^\mathrm{T})y\rangle =\overline{\mu}\langle x,y\rangle-\langle x,(A-A^\mathrm{T})y\rangle. $$ Rearranging and applying standard inequalities then yields $$ |\langle x,y\rangle| \leq\frac{\|A-A^\mathrm{T}\|_2}{|\lambda-\overline{\mu}|}. $$ The numerator of this fraction makes intuitive sense, since any correlation between eigenvectors must come from the skew-symmetric part of the matrix. Furthermore, the denominator provides the relationship you suggested in your question, especially when the eigenvalues are real.

share|improve this answer

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.