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Question is

Substitute $A=SBS^{-1}$ into the product $$ (A-y_1I)(A-y_2I)\cdots(A-y_nI) $$ and show that the product equals $0$

where $B$ denotes the diagonal form of $A$ with eigenvalues on the diagonal and $y$ denotes the eigenvalues

Question says that this is related to the Cayley Hamilton Theorem. I know that Cayley Hamilton Theorem states that the eigenvalues in the characteristic polynomial of the matrix can be exchanged with the matrix itself. But I couldn't apply it here.

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Hint: notice that $I = S I S^{-1}$, so $A - y_1 I = SBS^{-1} - y_1 SIS^{-1} = S (B - y_1 I) S^{-1}$.

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  • $\begingroup$ Wow. Thanks for the hint, how would I notice that in any other different time ) : $\endgroup$ – Xenidia Aug 1 '16 at 14:27
  • $\begingroup$ @PatrickStevens But $S (B - y_1 I) S^{-1}$ implies that the given expression is zero? $\endgroup$ – user1942348 Aug 1 '16 at 16:16
  • $\begingroup$ @Xenidia The way I came up with it was to think what Cayley-Hamilton would have given me, and to think how I could transform it to the actual expression we are to consider. Also as soon as I see conjugation, I think "this is a homomorphism, I'm going to want the $S^{-1}$ of one component to cancel with the $S$ of the next". $\endgroup$ – Patrick Stevens Aug 1 '16 at 17:19
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    $\begingroup$ @user1942348 It's not necessarily true that $B-y_1I$ is equal to $0$, but it is true that $(B-y_1 I) (B-y_2 I) \dots (B-y_n I)$ is equal to $0$ (by Cayley-Hamilton). $\endgroup$ – Patrick Stevens Aug 1 '16 at 17:19

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