Vtyjkyuk

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 $

By assumption, $p(x) = x^2 +1$ nullifies $boldsymbol A$. Since $boldsymbol A$ is $2times 2$, 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:

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


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


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

Hint: Suppose $v$ is an eigenvector with eigenvalue $lambda$. We then have
$$ 0 = mathbf0v = (A^2+I)v = (lambda^2+1)v $$

For any eigenvalue, $lambda$, and the associated eigenvector, $v$, we have
$$
Av=lambda v
$$
which means
$$
-v=A^2v=lambda^2v
$$

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

$$sigma(A) subseteq textzeroes of m_A subseteq textzeroes of x^2 + 1 = i, -i$$

Hence $sigma(A) cap mathbbR = emptyset$.