# Show that $\det(A-\lambda I)=\det(B-\lambda I)$

Alright, I've been trying to work this linear algebra problem out for a bit and I don't seem to be getting anywhere. The problem is this:

Assume that $A=M^{-1}BM$. Show that $\det(A-\lambda I)=\det(B-\lambda I)$.

So my instincts tell me that this has something to do with the fact that $A^n$ can be expressed as $M^{-1}BM$

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This rather has something to do with the fact that det(UV)=det(U)det(V) for suitable matrices U and V... –  Did Oct 5 '11 at 7:20
Hint: $\lambda I = M^{-1}(\lambda I) M$. –  Ted Oct 5 '11 at 7:23
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## 1 Answer

A small hint: In the expression $\det(A-\lambda I)$ you can use the given fact that $A=M^{-1}BM$ of course, but also that $I=M^{-1}IM$.

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BAH! That was simple. I should not be doing these things at midnight. Thank you so much, problem solved. –  crf Oct 5 '11 at 7:27
@Colin: You're welcome. –  Hans Lundmark Oct 5 '11 at 7:32
@Colin: If you're happy with Hans's answer, please remember to click the checkmark to the left of his answer, so that the software considers this resolved. :) –  Ｊ. Ｍ. Oct 5 '11 at 8:59
Thanks, didn't know that. –  crf Oct 12 '11 at 5:35
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