if $\ A^2 = -I $ then $\ A $ has no real eigenvalues Given $\ A $ is a $\ 2 \times 2 $ matrix over $\ \mathbf R$ that $\ A^2 = -I $ and I need to prove $\ A $ has no real eigen values.
$\ A^2 = -I \rightarrow A^2 +I = 0 $
I guess it is something about $\ x^2 +1 = 0 $ has no real solutions... but if someone can show me the connection to matrices (if it is about that?)  I mean how can I conclude anything from that about the characteristic polynomial of $\ A $

Since $A^2+I = 0$, the polynomial $x^2+1$ annihilates the matrix $A$. Therefore the minimal polynomial $m_A$ of $A$ divides $x^2 + 1$ so
$$\sigma(A) \subseteq \{\text{zeroes of } m_A\} \subseteq \{\text{zeroes of } x^2 + 1\} = \{i, -i\}$$
Hence $\sigma(A) \cap \mathbb{R} = \emptyset$.
