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.

Let $A\in\mathbb{R}^{n\times n}$ denote some symmetric, and $B\in\mathbb{R}^{n\times n}$ some positive-definite matrix. The generalized eigenvalue problem, $[A, B]$, corresponds to a scalar-vector pair, $(\lambda, u)$, satisfying $$Au=\lambda Bu.$$

What is the property of generalized eigenvectors $u$, e.g., are they mutually orthogonal? Are they somehow related to eigenvectors of $A$ (or $B$ )?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

If $B$ is invertible, then you can rewrite this equation as $B^{-1}Au=\lambda u$, so you get an ordinary eigenvector equation, and thus you get all the properties of normal eigenvectors.

Moreover, if $A$ is invertible, you can rewite the equation as $\lambda^{-1} u = A^{-1}Bu$ to again get an ordinary eigenvalue problem (however note that in this case, the vectors from $B$'s null space cannot be generalized eigenvalues, despite being eigenvalues of $A^{-1}B$, because $0$ has no inverse).

The interesting case of course is is when $B$ is not invertible. Obviously if the intersection of the null spaces of $A$ and $B$ doesn't vanish, every vector from that intersection is a generalized eigenvector to an arbitrary generalized eigenvalue (because then the equation reduces to $0=0$ irrespective of $\lambda$). Moreover, if $v_0$ is in the intersection of both null spaces, and $v$ is a generalised eigenvector with generalized eigenvalue $\lambda$, then $v+v_0$ is, too.

If $v$ is in $A$'s null space, but not in $B$'s, then it is a generalized eigenvector to the generalized eigenvalue $0$. If $v$ is in $B$'s null space but not in $A$'s, then it cannot be a generalized eigenvalue.

I guess for vectors $v\perp \operatorname{Ker}(B)$ the problem can again be reduced to an ordinary eigenvalue problem, but I don't currently see how.

share|improve this answer
    
I stated that $B$ is a positive definite matrix, hence invertible. You state that the generalized eigenvectors are eigenvectors of $B^{-1}A$, hence mutually orthogonal. However, on some places I read the the generalized eigenvectors are $B-$orthonormal, i.e., $u_i^TBu_j=0$, for $i\neq j$, and that they are $B-$normalized, i.e., $u_i^TBu_i=1$. Is this formulation a degree of freedom, i.e., the generalized eigenvectors can be expressed that way? –  user506901 Aug 28 '12 at 13:17
    
I suppose I make wrong assumption that the eigenvectors of each matrix are orthogonal; while this is the case with symmetric matrices, $B^{-1}A$ is generally not symmetric. –  user506901 Aug 28 '12 at 13:43
1  
Ah, I overlooked that B was positive definite. But then, with that restriction my answer could simply be reduced to the first sentence. And indeed, the eigenvectors need not be orthogonal; for example the eigenvectors of $\pmatrix{1&1\\0&2}$ are $\pmatrix{1\\0}$ and $\pmatrix{1\\1}$ which are clearly not orthogonal (at least not using the usual Euclidean scalar product). –  celtschk Aug 28 '12 at 15:04

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.