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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 $

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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.
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  • $\begingroup$ thanks for the elaborated answer! $\endgroup$ – bm1125 Sep 2 '18 at 9:27
  • $\begingroup$ Glad to help here! $\endgroup$ – xbh Sep 2 '18 at 9:28
  • $\begingroup$ Would you mind explaning what is the implications of a matrix without real eigenvalues? I mean does it mean anything about the determinant maybe? $\endgroup$ – bm1125 Sep 2 '18 at 9:33
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    $\begingroup$ Seems not so much consequences about determinants, since the constant term of the characteristic polynomial only possibly differentiate from the determinant a sign. $\endgroup$ – xbh Sep 2 '18 at 9:41
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Hint: Suppose $v$ is an eigenvector with eigenvalue $\lambda$. We then have $$ 0 = \mathbf0v = (A^2+I)v = (\lambda^2+1)v $$

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  • $\begingroup$ 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 $ ? $\endgroup$ – bm1125 Sep 2 '18 at 9:07
  • $\begingroup$ @bm1125: Edited with a different hint that is actually correct $\endgroup$ – Henning Makholm Sep 2 '18 at 9:11
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For any eigenvalue, $\lambda$, and the associated eigenvector, $v$, we have $$ Av=\lambda v $$ which means $$ -v=A^2v=\lambda^2v $$

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

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