# 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 $\ A$

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:

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.
• thanks for the elaborated answer! – bm1125 Sep 2 '18 at 9:27
• Glad to help here! – xbh Sep 2 '18 at 9:28
• Would you mind explaning what is the implications of a matrix without real eigenvalues? I mean does it mean anything about the determinant maybe? – bm1125 Sep 2 '18 at 9:33
• Seems not so much consequences about determinants, since the constant term of the characteristic polynomial only possibly differentiate from the determinant a sign. – xbh Sep 2 '18 at 9:41

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

• Do you mean that if $\ x^2 + 1$ is not the actual polynomial of $\ A$ it should be at least a root of the characteristic polynomial of $\ A$ ? – bm1125 Sep 2 '18 at 9:07
• @bm1125: Edited with a different hint that is actually correct – Henning Makholm Sep 2 '18 at 9:11

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

• If $\ \lambda$ is eigen value of $\ A$ then $\ \lambda^2$ is eigenvalue of $\ A^2$ ? – bm1125 Sep 2 '18 at 9:19
• Oh right, definition would be the quickest way. Thanks! – xbh Sep 2 '18 at 9:24
• @bm1125: indeed, with the same eigenvector: $A^2v=A(\lambda v)=\lambda Av=\lambda^2v$. – robjohn Sep 2 '18 at 9:38

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 $\sigma(A) \cap \mathbb{R} = \emptyset$.