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I've been trying to prove or disprove the following statement:

Let $A$ be a square matrix such that $\mathrm{rank}(A)=3$. Prove or disprove that if the characteristic polynomial of $A$ is $x^2(x-1)(x-2)$, then A is diagonalizable.

So I can see that the size of $A$ must be $4\times 4$. However, the fact that the rank of $A$ is $3$ made it almost impossible for me to find a (really) specific matrix, such that its characteristic polynomial is the one requested above. I couldn't find a counterexample, either.

Thanks!

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The statement is false. For instance, the matrix $$ A = \pmatrix{0&1&0&0\\0&0&0&0\\0&0&1&0\\0&0&0&2} $$ Has the appropriate rank and eigenvalues, but fails to be diagonalizable.

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What about

$$A= \begin{pmatrix} 0&1&0&0\\ 0&0&0&0\\ 0&0&1&0\\ 0&0&0&2 \end{pmatrix}$$

Which is not diagonalizable as its minimal polynomial do not have only simple roots.

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If $A$ is $n\times n$ and diagonalizable, then $x^m\mid \det(A-xI)$ if and only if $\operatorname{rk}A\le n-m$.

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