Show that there is no $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, but there is $B\in M_2(\mathbb{R})$ and $C\in M_3(\mathbb{C})$ whose minimal polynomials are both $x^2+1$. $M_n(\mathbb{F})$ denotes the vector space of $n\times n$ matrices over the field $\mathbb{F}$.
My work:
So if there exists $A\in M_3(\mathbb{R})$ whose minimal polynomial is $x^2+1$, its characteristic polynomial must be of the form $(x-a)(x^2+1)$ for some $a$, since the minimal polynomial divides the characteristic polynomial and the characteristic polynomial has degree $3$. I'm not able to find a contradiction from here.
For $B\in M_2(\mathbb{R})$, by a similar argument, $x^2+1$ must be $B$'s characteristic polynomial. Another thing I noticed is that $B$ cannot be diagonalizable, since if it were, $x^2+1$ should have split into distinct linear factors over $\mathbb{R}$, but that is clearly not the case. Doing a brute force calculation, considering $$B = \left[\begin{matrix} a & b \\ c & d \end{matrix}\right]$$ we get $x^2 – (a+d)x + ad – bc = x^2 + 1$, so $a+d = 0$ and $ad-bc=1$. This gives $a^2+1 = -bc$. Choosing $(a,b,c,d) = (1,-2,1,-1)$ yields a desirable $B$.
I am planning on doing something similar for $C\in M_3(\mathbb{C})$, but seems like it would be overkill. Is there an easier way around this? It'd be great if someone could help me out with disproving the existence of $A$, proving the existence of $C$, and finding a somewhat better solution for the existence of $B$.
Thanks!
Best Answer
Hint: Consider these:
$A\in M_3(\mathbb{R}) \implies \det A \in\mathbb{R}$
$A^2=-I \implies (\det A)^2 = \det (A^2) = \det(-I) = -1$
The simplest $B$ is $ \begin{pmatrix}0&-1\\1&\hphantom-0\end{pmatrix} $ (see also Matrix representation of complex numbers).
The simplest $C$ is $ \begin{pmatrix}i&0&0\\0&i&0\\0&0&-i\end{pmatrix} $