Can we prove that the eigenvalues of an $n\times n$ orthogonal matrix are $\pm 1$ from the definition of orthogonal matrix alone?

An $n\times n$ matrix $A$ is orthogonal iff $AA^T=A^TA=I$.

Is it possible to use solely this definition, without using the result proven from this definition which says orthogonality preserves length, to show that the eigenvalues of $A$ is $\pm 1$?

  • $\begingroup$ The fact that an orthogonal matrix preserves length is a theorem which uses this definition. $\endgroup$ – user46372819 Dec 11 '17 at 22:44
  • $\begingroup$ @Gibbs You have prove that. It's probably obvious to you because it is an easy proof. But still, you have to prove it. $\endgroup$ – user46372819 Dec 11 '17 at 22:53

Suppose $x$ is an eigenvector with corresponding eigenvalue $\lambda$. Then $x^T A^T Ax = (Ax) \cdot (Ax) = |Ax|^2 = \lambda^2 |x|^2$. But you also know that $x^T A^T A x = x \cdot x = |x|^2$. So $\lambda^2 = 1$ i.e. $\lambda = 1,-1$. Note this says that if the matrix is diagonalizable then the eigenvalues must be $1$ and/or $-1$. This doesn’t in and of itself show that an orthogonal matrix is always diagonalizable (indeed it doesn’t have to be over the reals).

  • $\begingroup$ Since the characteristic equation is a real cubic, it always has at least one real root, which implies that $A$ does always have a real eigenvector, which by the above has eigenvalue $\pm 1$. $\endgroup$ – Joppy Dec 11 '17 at 22:49
  • $\begingroup$ @Joppy The characteristic polynomial has degree $n$; for $n$ odd there will be a real root; for $n$ even there need not be. $\endgroup$ – egreg Dec 11 '17 at 22:54
  • $\begingroup$ @egreg: Too true, I read the question as $3 \times 3$ instead of $n \times n$ for some reason. $\endgroup$ – Joppy Dec 11 '17 at 22:55
  • $\begingroup$ This basically reproves that $A$ preserves length: $$\|Ax\|^2 = \langle Ax, Ax \rangle = x^TA^TAx = x^Tx = \langle x, x \rangle = \|x\|^2$$ and then just uses it: $$\|x\|^2 = \langle Ax, Ax \rangle = \langle \lambda x, \lambda x \rangle = |\lambda|^2 \|x\|^2 \implies |\lambda|^2 = 1$$ I believe the idea was to circumvent the inner product entirely. $\endgroup$ – mechanodroid Dec 11 '17 at 23:05

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