How to prove that if $A,B,M\in \mathcal{M_n}(\mathbb{C})$ and $\lambda,\mu\in\mathbb{C}$ so $$ \begin{cases} M&=& \lambda A+\mu B \\ M^2&=& \lambda^2 A+\mu^2 B \\ M^3&=& \lambda^3 A +\mu^3 B \end{cases} \Rightarrow M\ \text{is diagonalizable}.$$
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1What have you tried? Is this homework? Please rewrite it, so that it is a question, not a command. – Julian Kuelshammer Feb 21 '13 at 12:32
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1What do the rectangles mean? – user1551 Feb 21 '13 at 13:00
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2This question is just like some of Jules Verne's great novels: dark and misterious... – DonAntonio Feb 21 '13 at 13:27
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It does look like homework but I'll help you. Consider the polynomial $P(X)=X(X-\lambda)(X-\mu)$. You then have $P(M)=0$.
Case 1: $\lambda\mu\neq 0$ and $\lambda\neq \mu$. So the elementary divisors of $M$ are linear, hence $M$ is diagonalizable.
Case 2: $\lambda\mu = 0$, WLOG say $\lambda =0$, if $\mu =0$ you are done, otherwise $B^2=B$, hence $B$ is diagonalizable.
Case 3: $\lambda = \mu \neq 0$, as in case 2, $M$ is diagonalizable because $A+B = (A+B)^2$.
Nathan Portland
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