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If there is $n \times n$ matrix $A$, and $A^2= - I$?

Can anyone explain me why there exists no real eigenvalues? and $\det A=1$?

Przemysław Scherwentke
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dad
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  • This is pretty difficult to understand. Are you trying to say that $A = A^2 = -I$? – Cameron L. Williams Oct 30 '14 at 23:46
  • You should have precised that $A$ is a matrix with real coefficients. – Olórin Oct 30 '14 at 23:50
  • @MisesEnForce I thin that the question is about any matrix, as the second part leads to the real case. – Przemysław Scherwentke Oct 30 '14 at 23:53
  • @Przemysław Scherwentke If the matrix is allowed to be complex, then the question is irrelevant. Clearly the OP assumes (without writing it) that $A$ has real coefficients. – Olórin Oct 30 '14 at 23:57
  • See also: https://math.stackexchange.com/questions/471445/prove-properties-of-a2-i, https://math.stackexchange.com/questions/2007643/let-a-be-an-n-times-n-matrix-with-real-entries-such-that-a2-i-0-then, https://math.stackexchange.com/questions/949189/a-is-n-times-n-real-matrix-with-a2-i-to-prove-that-n-is-even, https://math.stackexchange.com/questions/2226327/an-especial-matrix-is-similar-to-an-especial-matrix – mdcq Apr 10 '18 at 16:34

2 Answers2

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I assume that $A$ is a matrix with real coefficients. Let $\lambda$ be a real eigenvalue and $X\in\mathbf{R}^n$, $X\not=0$ an eigenvector : $A X = \lambda X$. Then $-X = A^2 X = A ( A X ) = A (\lambda X) = \lambda A X = \lambda^2 X$, and then $\lambda^2 = -1$ as $X\not = 0$, which is impossible as $\lambda$ is real

Olórin
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Assuming you mean given a matrix $A$ such that $A^2 = -I$ show there are no real eigenvalues, here is how: Let $e$ be an eighenvalue of $A$, and ${\bf v}$ its corresponding eigenvector. Then $$ A{\bf v} = e{\bf v} $$ $$ -I{\bf v}=A^2{\bf v} = A(A{\bf v}e{\bf v}) = A(e{\bf v}) = e(A{\bf v}) = e^2{\bf v}$$ So $e^2 = -1$ and the egenvector is not real.

Mark Fischler
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