Find matrix $A$ with real entries with minimal polynomial $t^4 + 1.$

Question

The title says it all. This polynomial splits over $\mathbb{C}$ pretty obviously, $(t - \sqrt{i})(t + \sqrt{i})(t - i\sqrt{i})(t + i\sqrt{i}),$ but the matrix needs real entries, so I can't just make a diagonal matrix.

So my next idea is that $A^4 = -I,$ write $A^4 = B^2,$ where $B = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}.$ Then all I need to do is find $A$ such that $A^2 = B.$ Not possible (I think), probably because I started working with $2 \times 2$ matrices whereas the question does not give a size to he matrix, it just says "find a matrix with minimal polynomial ..."

This leaves me with questioning my approach, and that is why I've posted this question, how does one go about constructing a matrix given that it needs to satisfy a minimal polynomial, restricted to having elements in some field (preferably $\mathbb{R}$ or $\mathbb{C}$)? Solutions would be appriciated, but I am mostly interested in the procedure for constructing such matrices given a general minimal polynomial.

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For fun, the subquestion associated with the original question is, "Show that the usual real linear map $v \to Av$ has no non-trivial subspace."

Answer 1

Hint to find such a matrix:

1. decompose $t^4+1$ into quadratic/linear factors. Here, we have two quadratic factors $p_1(t)$ and $p_2(t)$.

2. for each factor $p_i(t)$, find a $2\times 2$ matrix $A_i$ with minimal polynomial $p_i$.

3. $A=\left(\begin{array}{ll}A_1&0\\0&A_2\end{array}\right)$.

Note:

1. Since each $p_i$ is irreducible, $A_i$ can be taken in the form $$\left(\begin{array}{rr} a& -b\\b&a\end{array}\right),$$ i.e, a rotation in $\mathbb{R}^2$. Like in Jack's answer, you should be able to find them both.
2. Clearly, $A$ has two stable subspaces :-).
3. $A$ should not have an invariant space, since each would yield an eigenvector hence an eigenvalue. But the minimal polynomial of $A$ has no real roots.

Answer 2

Put $$A = \left(\begin{matrix} 0&1&0&0\\ 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0 \end{matrix} \right).$$ Then $$A^2 = \left(\begin{matrix} 0&0&1&0\\ 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0 \end{matrix} \right), \,\,\,\,\, A^3 = \left(\begin{matrix} 0&0&0&1\\ -1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0 \end{matrix} \right), \,\,\,\,\, A^4 = \left(\begin{matrix} -1&0&0&0\\ 0&-1&0&0\\ 0&0&-1&0\\ 0&0&0&-1 \end{matrix} \right). $$ Thus $A^4 + I = 0$. Further, $A^3, A^2, A, I$ are linearly independent, so there is no third degree polynomial which annihilates $A$. Thus $t^4 + 1$ is the minimal polynomial of $A$.

I'm led to believe that if $A$ is $4\times 4$, the second part isn't necessarily true and we can only prove that $A$ has no $1$ or $3$ dimensional invariant subspaces while there are cases where $A$ has a $2$ dimensional invariant subspace. That being said, I'm struggling to find an example.

As an aside, did you find this on the UCLA Basic Exam? I remember this question came up on one of those.

EDIT: I had posted a counterexample to the fact that $A$ has no non-trivial invariant subspaces, but Quang Hoang pointed out in the comments that my counterexample was incorrect.

Answer 3

Figured it out, thanks to @Robert Israel's comment to look for companion matrices:

The answer to this specific question is $A = \begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0 \end{pmatrix}$.
Thanks for the responses everybody.