Related Eigenproblems

Since the HEP is one of the best understood eigenproblems, it is helpful to recognize when other eigenproblems can be converted to it.

1. If $A$ is non-Hermitian, but $\hat{A}= \alpha SAS^{-1}$ is Hermitian for easily determined $\alpha$ and $S$, it may be advisable to compute the eigenvalues $\hat{\lambda}$ and eigenvectors $\hat{x}$ of $\hat{A}$. One can convert these to eigenvalues $\lambda$ and eigenvectors $x$ of $A$ via $\lambda = \hat{\lambda}/ \alpha$ and $x = S^{-1} \hat{x}$. For example, multiplying a skew-Hermitian matrix $A$ (i.e., $A^* = -A$) by the constant $\sqrt{-1}$ makes it Hermitian. See §2.5 for further discussion.

2. If $A = B^*B$ for some rectangular matrix $B$, then the eigenproblem for $A$ is equivalent to the SVD of $B$, discussed in §2.4. Suppose $B$ is $m$ by $n$, so $A$ is $n$ by $n$. Generally speaking, if $A$ is about as small or smaller than $B$ ($n \leq m$, or just a little bigger), the eigenproblem for $A$ is usually cheaper than the SVD of $B$. But it may be less accurate to compute the small eigenvalues of $A$ than the small singular values of $B$. See §2.4 for further discussion.

3. If one has the generalized HEP $Ax = \lambda Bx$, where $A$ and $B$ are Hermitian and $B$ is positive definite, it can be converted to a Hermitian eigenproblem 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 $Ax = \lambda Bx$. See §2.3 for further discussion.