TITLE: Examples of real $2\times2$ and complex $3\times3$ matrices with minimal polynomial $t^2+1$
QUESTION [4 upvotes]: 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?

REPLY [4 votes]: $$
\left(
\begin{array}{cc}
0 & 1 \\
-1 & 0
\end{array}
\right)
$$
$$ $$
$$
\left(
\begin{array}{ccc}
i & 0 & 0 \\
0 & i & 1 \\
0 & 0 & i
\end{array}
\right)
$$

REPLY [2 votes]: 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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