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Let $A\in \text{Mat}_{3\times3}(\mathbb{R})$ such that $A^2-2A+I=0$ and $A\neq I$.

Find the Jordan form of $A$.

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You can factorise your equation as $$ (A-I)^2 = 0, $$ and hence all of $A$'s eigenvalues are $1$. Further, the minimal polynomial of $A$ is not $t-1$, because $A \neq I$, and hence the minimal polynomial is $(t-1)^2$. Hence you need to look at the possible Jordan blocks, and find the one that squares to the zero matrix. In particular, $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}^2 = 0, $$ but $$ \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}^2 \neq 0. $$

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Since $A^2 - 2A + I = 0$, the minimal polynomial $m_A(x)$ of $A$ must divide $x^2 - 2x + 1 = (x - 1)^2$. Since $A \neq I$, we must have $m_A(x) = (x - 1)^2$. Since the minimal polynomial must have the same linear factors as the characteristic polynomial, we see that the characteristic polynomial $p_A(x)$ must be $p_A(x) = (x - 1)^3$ so the only eigenvalue $A$ has is $1$. In your case, this gives you enough info to determine the Jordan form.

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