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Consider the following corrolary (3.3.4 in Horn's and Johnson's Matrix Analysis book):

For each $A\in\mathbb{C}^{n\times n}$, the minimal polynomial $q_A(t)$ divides the characteristic polynomial $p_A(t)$. Moreover, $q_A(\lambda)=0$ if and only if $\lambda$ is an eigenvalue of $A$, so every root of $p_A(t)=0$ is a root of $q_A(t)=0$.

There is no real $3\times3$ matrix with minimal polynomial $t^2+1$ (proofs for that are here). However, there is a real $2\times2$ matrix and a complex $3\times3$ matrix with minimal polynomial $q_A(t)=t^2+1$. I am trying to find such examples.

For the $2\times2$ case consider $$A=\begin{bmatrix}1&-1\\2&-1\end{bmatrix}$$ Its characteristic polynomial is $$p_A(t)=\det(tI-A)=\dots=t^2+1=(t-i)(t+i)$$ Then, by the above theorem and by the definition of the minimal polynomial (the unique monic polynomial of minimum degree, $q_A(t)$, that annihilates $A$ i.e. $q_A(A)=0$) we have $$q_A(t)|p_A(t)$$ and every eigenvalue is a root of $q_A(t)$. Thus, the minimal polynomial is indeed $q_A(t)=t^2+1$.

For the $3\times3$ case consider $$A=\begin{bmatrix}i&0&0\\0&1&-1\\0&2&-1\end{bmatrix}$$ Similarly, it is easy to see that $$p_A(t)=\det(tI-A)=\dots=(t-i)^2(t+i)$$ and among the two possibilities for the minimal polynomial $$p_1(t)=(t-i)^2(t+i)\qquad\text{and}\qquad p_2(t)=(t-i)(t+i)$$ we see that $p_2(t)=t^2+1$ is the one with minimum degree that annihilates $A$ so that $q_A(t)=t^2+1$.

Are the above examples correct? Any other examples, perhaps more interesting?

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    $\begingroup$ Yes, the examples are correct. I don't know what more exciting you are expecting. $\endgroup$ – Dietrich Burde Mar 1 '18 at 19:16
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$$ \left( \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right) $$ $$ $$ $$ \left( \begin{array}{ccc} i & 0 & 0 \\ 0 & i & 1 \\ 0 & 0 & i \end{array} \right) $$

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For $2\times 2$ matrices we know that the characteristic polynomial of $A$ is $\chi_A(x)=x^2-Tr(A)x+det(A)$ so if you want it to be $x^2+1$ then $Tr(A)=0$ and $det(A)=1$ so you can find any other example you wish.

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