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Suppose that $A$ is a Hermitian matrix and that $u,v$ are two vectors. Is there some known function $\kappa(A)$ so that $||u-v|| \leq \kappa(A) |\frac{u^{\*}Au}{u^{\*}u}-\frac{v^{\*}Av}{v^{\*}v}|$?

UPDATE: Andrew T. Baker has shown that the answer is "no" in general (to take an even simpler counter-example, take $A=I$) - so let's add the assumption that $u$ is a simple eigenvector of $A$.

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minor observation: on the rhs you don't need the norm, it's just scalar quantities. –  Suvrit Aug 5 '12 at 20:35
    
thanks! any ideas about the answer? :) –  Felix Goldberg Aug 6 '12 at 21:01

3 Answers 3

up vote 3 down vote accepted

I think the answer for a general Hermitian matrix $ A $ is no.

Let $ u, v $ be distinct eigenvectors of $ A $ with the same eigenvalue $ \lambda $ and normalized so that $ u^\ast u = v^\ast v = 1 $. Then $ \| u - v \| > 0 $ and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda u^\ast u - \lambda v^\ast v | = | \lambda - \lambda | = 0. \end{equation*}

A concrete counterexample is \begin{equation*} A = \left( \begin{array}{ccc} 2& & \\\\ & 2& \\\\ & & 1 \end{array} \right), u = \left( \begin{array}{c} 1 \\\\ 0 \\\\ 0 \end{array} \right), v = \left( \begin{array}{c} 0 \\\\ 1 \\\\ 0 \end{array} \right) \end{equation*}

ADDED: if we assume all eigenvalues are distinct, then the above argument leads to some kind of bound. Again, $u,v$ are eigenvectors of $ A $ with corresponding eigenvalues $ \lambda_1, \lambda_2 $ and normalized with $ u^\ast u = v^\ast v = 1 $. Then \begin{equation*} \| u - v \| \leq 2 \end{equation*} and \begin{equation*} | u^\ast Au - v^\ast A v | = | \lambda_1 - \lambda_2 | \end{equation*}

so that if we choose \begin{equation*} \kappa(A) \geq \min_{\lambda_i, \lambda_j} \frac{2}{|\lambda_i - \lambda_j|} \end{equation*} we can guarantee your inequality will hold.

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Thanks. I'll edit the question to focus on the case that $u$ is a simple eigenvector - which is the one that I'm basically interested in. –  Felix Goldberg Aug 7 '12 at 15:55
    
I've edited my answer to try to address this case, in which case you can get some kind of expression if you assume $ u,v $ normalized. But based on the answer of Sharkos, that normalization is necessary. –  Andrew T. Barker Aug 7 '12 at 16:41
    
But what if $v$ is not an eigenvector? I intend it to be free. –  Felix Goldberg Aug 7 '12 at 17:04
    
We have already assumed $A$ is diagonalizable, so $v$ can be written as a linear combination of eigenvectors, and everything else will follow pretty much as before. –  Andrew T. Barker Aug 7 '12 at 17:20

Another trivial problem, I'm afraid:

The two sides scale differently under rescaling of $u, v$. That is, if it holds for some $\kappa, u \neq v$ then simply multiply $u, v$ by some huge number $Z$ - then the left-hand side is now $Z$ times bigger whilst the right-hand side is constant.

How many further qualifications are necessary?!

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Well, yes, $u,v$ should be normalized in some way. Thanks for pointing that out. Is the question still trivial?... –  Felix Goldberg Aug 7 '12 at 17:05

If u is an eigenvector for an eigenvalue that is in the interior of the spectrum, then there is no such bound becomes there are numerous vectors that generate Rayleigh quotients for points in the convex hull of the spectrum. And as for Sharkos comment, you would want something like the sine of the angle between u and v.

An absolute condition number for a Rayleigh quotient would look more like $|x^*Ax-y'Ay|<=\kappa$(x'Ax) ||x-y||$, where K would be the condition number for x'Ax, where x and y are unit vectors. It shows how changes in the Rayleigh quotient can be bounded by changes in the vector.

Your suggested condition number is more like the condition number for the generating vector.

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