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I came across a question:

Prove or disprove: $A$ is a square matrix of size $n$ over $\mathbb R$, $P_{(\lambda)}=\lambda^k-1,P_{(A)}=0, k\geq 1 \implies A$ is diagonalizable over $\mathbb R$

So the obvious example is $A=I$ where you get that $I^3=I$ but i'm not sure if there are any other matrices over $\mathbb R$ can be powered into $I$.

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    $\begingroup$ Any rotation matrix with angle a rational multiple of π has the property $A^k=I$ for some $k$. $\endgroup$
    – Bernard
    Jul 3, 2016 at 11:50

2 Answers 2

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An even simpler counter-example: $A= \begin{pmatrix}-1 & 0 \\ 0 & -1 \end{pmatrix}$.

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You only can say it is diagonalisable on $\mathbf C$, since it implies the minimal polynomial of $A$ has simple roots.

Counter-example on $\mathbf R$

Set $\;A=\begin{pmatrix}1&0&0\\0&-2&-3\\0&1&1\end{pmatrix}$. The characteristic polynomial of $A$ is $$\chi_A(\lambda)=(1-\lambda)\bigl((\lambda+2)(\lambda-1)+3\bigr)=-(\lambda-1)(\lambda^2+\lambda+1)=1-\lambda^3, $$ and by Hamilton-Cayley, $I-A^3=0$. However, the matrix is not diagonalisable on $\mathbf R$ since two of its eigenvalues are the complex third roots of unity.

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