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Given a Hermitian indefinite pencil $(A-\lambda B)$ where both $A=A^H$ and $B=B^H \in \mathbb{C}^{n\times n}$ are possibly indefinite, it is straightforward to show that the eigenvalues are either real or come in complex conjugate pairs. Similarly, given a real nonsymmetric pencil $(M-\lambda N)$ where both $M,N \in \mathbb{R}^{n\times n}$ are nonsymmetric, it is straightward to show that the eigenvalues are also either real or come in complex conjugate pairs.

My question now is, does there exist a unitary similarity transform to convert $(A-\lambda B)$ into a real nonsymmetric pencil computable in a finite number of steps?

There exist methods to convert $(A-\lambda B)$ into $(T-\lambda S)$ where $T$ is real tridiagonal and $S$ is real diagonal, but the transformation is not unitary. Note that for this question, there is no requirement of obtaining a condensed form for the real matrices.

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  • $\begingroup$ how about $\lambda=0$ ? I.e. can you do this for $A$ alone? $\endgroup$ – Dima Pasechnik Feb 8 '15 at 9:32

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