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 $
 A: Hint: Suppose $v$ is an eigenvector with eigenvalue $\lambda$. We then have
$$ 0 = \mathbf0v = (A^2+I)v = (\lambda^2+1)v $$
A: By assumption, $p(x) = x^2 +1$ nullifies $\boldsymbol A$. Since $\boldsymbol A$ is $2\times 2$, the characteristic polynomial of $\boldsymbol A$ must be $p(x)$. Hence $\boldsymbol A$ has no real eigenvalues since $p(x)=0$ has no real solutions. 
Concepts related: minimal polynomial, characteristic polynomial. 
Facts assumed in advance: 


*

*For $\boldsymbol A \in \mathrm M_n(\Bbb F)$, where $\Bbb F$ is a number field, the characteristic polynomial must be of degree $n$.

*Characteristic polynomials nullifies the matrix [Cayley-Hamilton theorem]. 

*Minimal polynomial divides every nullifying polynomial, including the characteristic polynomial. 

A: For any eigenvalue, $\lambda$, and the associated eigenvector, $v$, we have
$$
Av=\lambda v
$$
which means
$$
-v=A^2v=\lambda^2v
$$
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 $\sigma(A) \cap \mathbb{R} = \emptyset$.
