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I am looking for an elegant proof of the following: Let $A,B \in \mathbb{K}^{n \times n}$ be diagonalizable matrices, and $P_A, P_B$ their characteristic polynomials. Then:

$A, B$ similar $\Leftrightarrow$ $P_A = P_B$

I am somehow stuck at this.

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Hint: Matrices satisfy their characteristic polynomials. What happens when you substitute a matrix similar to $A$ into $A$'s characteristic polynomial?

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  • $\begingroup$ Very nice approach! However I can't think through it. Say A similar to B $\Rightarrow$ $A = S B S^{-1}$. Then I have $P_B(A) = \sum_{i=1}^n c_i A = \sum_{i=1}^n c_i S B S^{-1} = S \left( \sum_{i=1}^n c_i B \right) S^{-1} = S 0 S^{-1} = 0$. So i get $P_A(B) = P_B(A) = 0$. Using the minimal polynomials $M_A$ and $M_B$, I get $M_A|P_B$ and $M_B|P_A$. Also $P_A$ and $P_B$ have the same degree, but I can't see why herby follows $P_A=P_B$? $\endgroup$
    – Haatschii
    Oct 10 '12 at 21:04
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If $A$ and $B$ are similar, then $B=XAX^{-1}$. So $$ P_B(t)=\det B-t\,I=\det XAX^{-1}-t\,XX^{-1}=\det X(A-t\,I)X^{-1}=\det A-t\,I=P_A(t). $$

Conversely, if $A$ is diagonalizable, then $A=SDS^{-1}$, with $D$ diagonal. Then $$ P_A(t)=\det A-t \,I=\det SAS^{-1}-tSS^{-1}=\det S(D-t\,I)S^{-1}\\=\det D-t\,I=(D_{11}-t)\cdots(D_{nn}-t). $$ So, if $P_A=P_B$, we conclude that $A=SDS^{-1}$, $B=TDT^{-1}$ for the same diagonal matrix $D$, and thus $$ B=TDT^{-1}=TS^{-1}AST^{-1}=(TS^{-1})A(TS^{-1})^{-1}. $$

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