I am working on the following problem:
Find a $4 \times 4$ matrix over $\mathbb{Q}$ such that $A^4 = -I$.
I know that if $A^4 = -I$, it would suffice to find a $4 \times 4$ matrix over $\mathbb{Q}$ such that its characteristic polynomial is $p(x) = x^4 + 1$, since every matrix satisfies its own characteristic polynomial by the Cayley-Hamilton Theorem. In this case, it seems plausible to me that it would have to have main diagonal entries all $0$ in order to get the $x^4$ term showing up in $p(x)$. But when I try to vary what the off-diagonal entries are, then, I can't seem to get the appropriate characteristic polynomial. Is there a more clever way to approach this than to ad-hoc construct such a matrix?
Thanks!
As in my comment, you can take the matrix $\begin{pmatrix}0&0&0&-1\\1&0&0&0\\0&1&0&0\\0&0&1&0\end{pmatrix}$. You may check that this has the desired characteristic and minimal polynomial or just note that with $\{e_i\}$ the standard basis that this maps $e_1\to e_2\to e_3\to e_4\to -e_4$ so that it will have the desired property.