Eigenvalues $\lambda$ of a matrix $A$ are defined as $Ax = \lambda x$, where $x$ are the eigenvectors. The characteristic polynomial is $\det(A-\lambda I)$.

Say I have the expression $Ax = \lambda Bx$, where $B$ is a matrix as well. Does this equation have a characteristic polynomial that enables me to find its eigenvalue $\lambda$?

  • $\begingroup$ @moose thanks, I edited it $\endgroup$ – BillyJean Jul 24 '13 at 7:29
  • $\begingroup$ Something like $\operatorname{det}(A - \lambda B)$? $\endgroup$ – Elchanan Solomon Jul 24 '13 at 7:33
  • $\begingroup$ yes, polynimoial, not equation. Sorry. @IsaacSolomon That is also what I thought, but I was unsure if I am allowed to make that generalization $\endgroup$ – BillyJean Jul 24 '13 at 7:36
  • $\begingroup$ @GitGud Thanks, done $\endgroup$ – BillyJean Jul 24 '13 at 7:37

You are always allowed to do the generalization proposed by Isaac, which remains valid, even if $B$ is not invertible.

Namely, there is a non-zero vector $x$ such that

$$ Ax = \lambda Bx \qquad \Longleftrightarrow \qquad (A - \lambda B)x = 0 \ . $$

Which means that the matrix $A - \lambda B$ has a non-trivial null space (for instance, $x$ belongs to it). Which means its determinant must be zero and vice-versa: if $\det (A-\lambda B) = 0$, then the matrix $A - \lambda B$ has non-trivial null space.

All in all,

$$ \text{There is $\ x \neq 0\ $ such that} \ \ Ax = \lambda Bx \qquad \Longleftrightarrow \qquad \det (A - \lambda B) = 0 \ . $$

And I guess you can call $\det (A - \lambda B)$ the characteristic polynomial of $A$ and $B$, or something like that if you whish.

EDIT. If the matrix $B$ is invertible, then $\det (B) $ and $ \det (B^{-1}) = 1/\det(B) \neq 0$ and the last equality is equivalent to

$$ 0 = \det (B^{-1}) \det (A - \lambda B) = \det (B^{-1} A - \lambda B^{-1}B ) = \det (B^{-1} A - \lambda I ) \ . $$

  • $\begingroup$ But $\det(A-\lambda B)=0$ is different from $\det(B^{-1}A-\lambda)=0$, which is what Isaac showed? $\endgroup$ – BillyJean Jul 24 '13 at 8:09
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    $\begingroup$ Ok, I've added the case $B$ invertible, which happens to be just that: a particular one. $\endgroup$ – d.t. Jul 24 '13 at 8:14

Suppose B is invertible, so $B^{-1}$ exists. This means you can re-write your equation to:

$\begin{align}Ax &= \lambda B x\\ \Leftrightarrow Ax &= B \lambda x\\ \Leftrightarrow \underbrace{B^{-1} A}_{=: C} x &= \lambda x\\ \Leftrightarrow Cx &= \lambda x \end{align}$

So you end up in the same situation as before.


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