Suppose $A \in \mathcal{O}(n,\mathbb{R})$. Then $A^{T}A=AA^{T}=A^{*}A=AA^{*}=1$ (where * denotes the Hermitian conjugate). Thus $A$ is normal and hence by spectral theorem it has a decomposition such that $A=SDS^{*}$ where $S \in \textbf{U}(n,\textbf{C})$ with orthonormal eigenvectors as columns and $D$ is complex diagonal. Then $$ A^{T}A=(SDS^{*})^{T}SDS^{*}=\bar{S}DS^{T}SDS^{*}=I $$

Where $I$ is the identity as usual.

I'm stuck at this point. Typically I would want to use the fact that S is unitary to simplify the issue, but I don't see how I can get the eigenvalues out of this. I also tried another method and got the following:

Suppose $A$ is as before. Let $v$ be an eigenvector of $A$ with eigenvalue $\lambda$, i.e. $Av=\lambda v$. Then $$ (Av)^T Av=(\lambda v)^T \lambda v \implies v^T A^T Av=v^T \lambda^{2} v $$ hence by orthogonality of A, $$ v^T v=\lambda^{2} v^T v \implies \lambda^2=1 $$

Thus $\lambda=\pm 1$.

Now this is all fine and good, but my class notes suggest that $|\lambda|=1$ and that the eigenvalues are either real or complex conjugate pairs. How do I get the conjugate pairs condition? It doesn't seem apparent from either of these methods. Thanks for listening!

  • 2
    $\begingroup$ The eigenvalues are roots of the characteristic polynomial. Roots of real polynomials are either real, or in conjugate pairs, see here; or this duplicate. $\endgroup$ – Dietrich Burde Mar 1 '18 at 20:21
  • $\begingroup$ Okay then the proof should look something like given $A$ orthogonal, $\det(A-\lambda I) = 0$ then $\lambda=e^{i\alpha}, e^{-i\alpha}$? I still don't see how this comes about. $\endgroup$ – lee.dul.lee Mar 1 '18 at 20:27
  • $\begingroup$ Just focus on the polynomial, e.g., $\chi(t)=t(t^2+1)$. Either it has a real root, or it has complex conjugate ones, or both. This holds for all real polynomials, see the duplicate. So it has nothing to do with orthogonal matrices. Of course, the eigenvalues of $A$ satisfy $\lambda=\pm1$, which gives $|\lambda|=1$, right? $\endgroup$ – Dietrich Burde Mar 1 '18 at 20:33

If $A \in \mathcal{O}(n, \mathbb{R})$, then $A$ is also in $\mathcal{O}(n, \mathbb{C})$, and satisfies the property $\langle Av, Aw \rangle = \langle v, w \rangle$ for any two vectors $v, w \in \mathbb{C}^n$, where $\langle \cdot, \cdot \rangle$ is the standard inner product given on column vectors by $\langle v, w \rangle = \sum_i v_i \overline{w_i}$.

If $\lambda$ is an eigenvalue of $A$, then since we are over $\mathbb{C}$ it has an eigenvector $v$. Then $$\langle v, v \rangle = \langle Av, Av \rangle = \langle \lambda v, \lambda v \rangle = \lambda \overline{\lambda} \langle v, v \rangle$$ And since $\langle v, v \rangle \neq 0$, we have $|\lambda| = 1$. The fact that all eigenvalues are real or complex conjugate pairs comes from the fact that the characteristic polynomial has real coefficients.


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