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Basically what is given is the characteristic polynomial of A $$p(x)=(2-x)^2$$ and I am asked if the matrix A is diagonalizable. I have already realized that $x=2$ is a root (eigenvalue) with multiplicity 2 which implicates the matrix being diagonalizable if and only if the Eigenspace with respect to 2 has dimension 2 but I have no clue about how to find this matrix A and the eigenspace.

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    $\begingroup$ Without more information both cases can happen. If $A=\operatorname{diag}(2,2)$, then it is diagonalizable. If $A=\begin{pmatrix}2&1\\0&2\end{pmatrix}$, then it is not. $\endgroup$
    – logarithm
    Commented May 25, 2019 at 14:31
  • $\begingroup$ as @logarithm points out with his two examples, the purpose of this question is to show you that (in general) the characteristic polynomial alone doesn't give you enough information about diagonalizability $\endgroup$
    – peek-a-boo
    Commented May 25, 2019 at 14:38
  • $\begingroup$ I suspect that your question is something along the lines of: "Is $A$ necessarily diagonalizable?" This means that you're to either conclude (and prove) that knowing the characteristic polynomial is sufficient for determining diagonalizability, or conclude (as you have already done) that it isn't sufficient and demonstrate that by examples. $\endgroup$ Commented May 25, 2019 at 14:47

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Guide:

You just have to exhibit two matrices, one of which it is diagonalizable and one of which it is not.

You might want to try matrix of this form. $$\begin{bmatrix} 2 & x \\ 0 & 2\end{bmatrix}$$

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