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AntiElephant
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Homework Statement
If A and B are invertible matrices over an algebraically closed field [itex] k [/itex], show there exists [itex] \lambda \in k [/itex] such that [itex] det(\lambda A + B) = 0 [/itex].
The Attempt at a Solution
Can anyone agree with the following short proof? I tried looking online for a confirmation, but I wasn't exactly sure how to search exactly for this problem.
Suppose [itex] det(\lambda A + B) \neq 0 ~~\forall \lambda \in k [/itex]
Since [itex] det(A^{-1}) \neq 0 [/itex], then
[itex] det(A^{-1}(\lambda A + B)) = det(\lambda I + A^{-1}B) = det(A^{-1}B - (-\lambda)I) \neq 0 ~~\forall \lambda \in k [/itex]So [itex] -\lambda [/itex] is not an eigenvalue for [itex] A^{-1}B [/itex] for all [itex] -\lambda \in k [/itex]. i.e. [itex] \lambda [/itex] is not an eigenvalue of [itex] A^{-1}B [/itex] for all [itex] \lambda \in k [/itex]. Since [itex] k [/itex] is algebraically closed this cannot be the case.
Is this okay? I guess it is not necessarily true when [itex]k[/itex] is not algebraically closed?
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