Related Eigenproblems

1. If $A$ and $B$ are Hermitian, $B$ is not positive definite, but $\alpha A + \beta B$ is positive definite for some choice of real numbers $\alpha$ and $\beta$, one can solve the generalized Hermitian eigenproblem $A - \lambda (\alpha A + \beta B)$ instead. Let $\hat{B}= \alpha A + \beta B$; then the eigenvectors of $A - \lambda B$ and $A - \lambda \hat{B}$ are identical. The eigenvalues $\lambda_i$ of $A - \lambda B$ and the eigenvalues $\hat{\lambda}_i$ of $A - \lambda \hat{B}$ are related by $\lambda_i = \beta \hat{\lambda}_i / (1 - \alpha \hat{\lambda}_i)$.

2. If $A$ and $B$ are non-Hermitian, but $\hat{A}= \alpha SAT$ and $\hat{B}= \beta SBT$ are Hermitian, with $\hat{B}$ positive definite, for easily determined $\alpha$, $\beta$ and nonsingular $S$ and $T$, then one can compute the eigenvalues $\hat{\lambda}$ and eigenvectors $\hat{x}$ of $\hat{A}- \lambda \hat{B}$. One can convert these to eigenvalues $\lambda$ and eigenvectors $x$ of $A$ via $\lambda = \hat{\lambda}\beta / \alpha$ and $x = T \hat{x}$. For example, if $B$ is Hermitian positive definite but $A$ is skew-Hermitian (i.e., $A^* = -A$), then $\sqrt{-1}A$ is Hermitian, so we may choose $\alpha = \sqrt{-1}$, $\beta=1$, and $S=T=I$. See §2.5 for further discussion.

3. If one has the GHEP $Ax = \lambda Bx$, where $A$ and $B$ are Hermitian and $B$ is positive definite, then it can be converted to a HEP as follows. First, factor $B = LL^*$, where $L$ is any nonsingular matrix (this is typically done using Cholesky factorization). Then solve the HEP for $\hat{A}= L^{-1}A L^{-\ast}$. The eigenvalues of $\hat{A}$ and $A - \lambda B$ are identical, and if $\hat{x}$ is an eigenvector of $\hat{A}$, then $x = L^{-\ast}\hat{x}$ satisfies $Ax = \lambda Bx$. Indeed, this is a standard algorithm for $A - \lambda B$.

4. If $A$ and $B$ are positive definite with $A = R^*R$ and $B = U^*U$ for some rectangular matrices $R$ and $U$, then the eigenproblem for $A - \lambda B$ is equivalent to the quotient singular value decomposition (QSVD) of $R$ and $U$, discussed in §2.4. The state of algorithms is such that it is probably better to try solving the eigenproblem for $A - \lambda B$ than computing the QSVD of $R$ and $U$.