Since A and B are diagonalizable to C so
PAP^-1 =C and so as B(QBQ-1= C)
Which is PAP^-1 = QBQ-1= C
2026-03-25 23:38:38.1774481918
If A and B are both diagonalizable to another matrix C, is A similar to B?
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The answer is yes.
$$PAP^{-1}=QBQ^{-1} \implies A=P^{-1}QBQ^{-1}P$$
$$ A= (P^{-1}Q)B(P^{-1}Q)^{-1}$$