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 $B$ )?


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$
    – user506901
    Aug 28, 2012 at 13:17
  • $\begingroup$ 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. $\endgroup$
    – user506901
    Aug 28, 2012 at 13:43
  • 1
    $\begingroup$ 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). $\endgroup$
    – celtschk
    Aug 28, 2012 at 15:04

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