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Consider a matrix pencil of quadratic form $F-λB$ with $B$ positive definite.

For which $λ$ the pencil $F-λB$ less or equal to $0$ (negative definite)?

Michael Hardy
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Apostolos K
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    Writing F-λB makes the minus sign look like a hyphen instead of a minus sign. I changed it to $F-λB$. ${}\qquad{}$ – Michael Hardy Aug 31 '15 at 16:44
  • AK, can you answer it if $B=I?$ – Will Jagy Aug 31 '15 at 18:16
  • i think the solution for this problem will be with the arguments in the book by F Gantmacher: MATRIX THEORY I, 1959, pp 317-326, Extremal properties of the characteristic values of a regular pencil of forms. – Apostolos K Aug 31 '15 at 19:02

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As you are talking about quadratic forms, I assume that $F$ is Hermitian. By Sylvester's law of inertia, $F-\lambda B$ is negative definite if and only if $B^{-1/2}FB^{-1/2}-\lambda I$ is negative definite. In turn, it is negative definite if and only if $\lambda > \rho(B^{-1/2}FB^{-1/2})=\rho(FB^{-1})$. Replace $>$ by $\ge$ when negative semidefiniteness is concerned.

user1551
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