Let $X\in\mathbb{R}^{n\times p}$ denote a matrix with $p$ linearly-independent columns, and let $L\in\mathbb{R}^{n\times n}$ denote a symmetric matrix. Furthermore, let $D\in\mathbb{R}^{n\times n}$ denote a diagonal matrix with positive-entries.

Now suppose that we are interested in computing solutions to the symmetric generalized-definite eigenproblem,

$$ A v = \lambda B v, $$

where the symmetric matrices $A$ and $B$ (where $B$ is also positive-definite) are implicitly given by the matrices $X$, $L$, and $D$ above, where $A = X^T L X$, and $B = X^T D X$.

I ran into a paper which claimed that we may solve an equivalent standard eigenvalue problem involving a modification of $X$. In particular, if we orthogonalize $X$ with respect to $D$, say $\tilde X^T D \tilde X = I$, then eigenpairs of

$$ \tilde X^T L \tilde X u = \lambda u $$

are also eigenpairs for the original generalized eigenproblem. How can we show this to be true?

  • $\begingroup$ Could you clarify what your assumptions on $A$ and $B$ are? For instance, is $B$ symmetric/Hermitian and positive-definite? $\endgroup$ Commented Oct 3, 2012 at 20:00
  • $\begingroup$ @JackPoulson Edited: $A$ is symmetric, and $B$ is positive definite. However, I'm interested in the specific problem: $A=X^TLX$ and $B=X^TDX$ $\endgroup$
    – usero
    Commented Oct 3, 2012 at 20:18
  • $\begingroup$ I am still not sure what you're getting at with your notation, would you mind providing a citation rather than saying "found in the literature"? Have you looked into the Cholesky-Wilkinson algorithm? It also reduces a problem of your form to a standard eigenvalue problem (though it should not be used if B is ill-conditioned). $\endgroup$ Commented Oct 3, 2012 at 23:23
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    $\begingroup$ No citation still? Part of what doesn't make sense is that your (1) equation is $p \times p$, and your original $Av=\lambda Bv$ equation is $n \times n$. It would really help clear things up if you said how $X$, $L$, and $D$ relate to $A$ and $B$. $\endgroup$ Commented Oct 4, 2012 at 14:27
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    $\begingroup$ I hate to sound like a broken record, but, no citation? $\endgroup$ Commented Oct 4, 2012 at 16:58

1 Answer 1


Let $X=\tilde X Z$, where $\tilde X^T D \tilde X=I$. Then we may rewrite

$$ (X^T L X)v = \lambda (X^T D X) v $$


$$ (Z^T \tilde X^T L \tilde X Z) v = \lambda (Z^T Z)v. $$

Now, put $u=Z v$, and premultiply both sides of the previous equation by $Z^{-T}$ to find

$$ \tilde X^T L \tilde X u = \lambda u. $$

It is worth noting that, contrary to your last sentence, the $u$ and $v$ are not the same, but are related through the equation $u=Z v$. Thus, after solving the standard eigenvalue problem with $\tilde X^T L \tilde X$, linear systems will need to be solved with $Z$ in order to find the original eigenvectors.

It is also important to notice that the fact that $D$ is positive is crucial to the ability to orthonormalize $X$ with respect to $D$. Consider the case where $D=-I$. Then requiring that $\tilde X^T D \tilde X = I$ is equivalent to requiring that $\tilde X^T \tilde X = -I$, which is not possible for real $\tilde X$ (nor in the complex case, when we replace transposes with Hermitian-transposes). When $D$ is positive, we may define

$$ Z = \sqrt{X^T D X}, $$

which will exist because $X^T D X$ will be positive-definite.

  • $\begingroup$ Thanks. Now it is clear why I do not want to point so publicly to mistakes of some works. $\endgroup$
    – usero
    Commented Oct 4, 2012 at 18:05
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    $\begingroup$ @usero No, it really is not "clear". If a paper is erroneous, it is much better if that is documented somewhere. If the paper is correct, it would greatly clarify the question. Successive paraphrasing of even simple statements has a tendency to change them, so even if you are happy with evaluating the paper based on Jack's answer here, it would still help the community if you included a citation. $\endgroup$
    – Jed Brown
    Commented Oct 4, 2012 at 19:07
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    $\begingroup$ @usero Jack's answer is great, but it would have been very simple to show that the claim is false by counterexample. (Simple exercise left to you.) Adding a citation would have greatly simplified his guess-work and editing of your post. $\endgroup$
    – Stefano M
    Commented Oct 4, 2012 at 21:58
  • $\begingroup$ The same answer, slightly different wording, is here math.stackexchange.com/a/206604/43193 Note that the same constraint on $D$ positive. $\endgroup$
    – adam W
    Commented Oct 21, 2012 at 13:53

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