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Why are Eigenvectors of an orthogonal matrix with respect to different eigenvalues orthogonal to one another?

I tried to find this question, if this is a duplicate post a link and I will cancel this one.

Also take an orthogonal matrix $A \in O(3)$ and the linear application associated with it $f: R^3 \rightarrow R^3$

Why is it that if $1$ is an eigenvector then $dim(V_1) = R^3$ and $A = I$ but if $1$ is not an eigenvector then $dim(V_2) $ is $2$ or $1$?

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3 Answers 3

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Suppose $Q$ is orthogonal, $Qx=\lambda_1x$,$Qy=\lambda_2y$, and $\lambda_1 \neq \lambda_2$. Then

$$x^*y=x^* I y = x^* Q^* Q y = (Qx)^* (Qy) = \overline{\lambda_1} \lambda_2 x^* y.$$

So either $\overline{\lambda_1} \lambda_2=1$ or $x^* y = 0$. In the latter case you are done. Try to prove that the former is impossible, given the assumptions above.

Note that this also follows from the spectral theorem for normal matrices (a normal matrix $A$ satisfies $A^* A = A A^*$.)

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  • $\begingroup$ Thank you a lot Ian I will try to prove that $\bar{\lambda_1}\lambda_2 $ is different from one tomorrow. Could you also try to make sense of my question on the linear application? Is $V_1$ standard notation for something? Is it the matrix formed by all the eigenvectors? thanks again. $\endgroup$
    – Monolite
    Oct 14, 2015 at 21:55
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    $\begingroup$ @Monolite Sorry, I'm not familiar with the notation. $\endgroup$
    – Ian
    Oct 14, 2015 at 22:02
  • $\begingroup$ no you were right it does refer to the eigenspace! $\endgroup$
    – Monolite
    Oct 14, 2015 at 22:08
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    $\begingroup$ @Ian sorry to ask in a post that will soon be turning 7 years old, but I had a question: why can't we have $\overline{\lambda_1}\lambda_2=1$? What about $\lambda_1=k, \lambda_2=\frac1k$ with $k\in \mathbb{R}\setminus{\{0\}}$? $\endgroup$
    – Dr. Mathva
    Jul 3, 2022 at 21:43
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    $\begingroup$ @Dr.Mathva They have modulus 1 because $Q$ is orthogonal, and so if their product is 1 then they are conjugates of one another. But then $\lambda_1$ and $\lambda_2$ have to be the same, which we already assumed they're not. $\endgroup$
    – Ian
    Jul 4, 2022 at 1:50
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The linear function $f$ preserves the inner product, so if $v$ is an eigenvector with eigenvalue $\lambda$, then

\begin{eqnarray*} <v,v> &=& <f(v),f(v)> \\ &=& <\lambda v, \lambda v> \\ &=& \lambda^2 <v,v> \\ \end{eqnarray*}

Thus $\lambda = \pm 1$. Now if $v_1$ and $v_2$ are eigenvectors with eigenvalues $1$ and $-1$, respectively, then

\begin{eqnarray*} < v_1, v_2 > &=& < f(v_1), f(v_2) > \\ &=& < v_1, -v_2 > \\ &=& -<v_1,v_2> \\ \end{eqnarray*}

But this implies $<v_1,v_2>=0$, so $v_1$ and $v_2$ are orthogonal.

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    $\begingroup$ $\lambda^2=1$ doesn't necessarily mean $\lambda=\pm 1$. Eigenvalues can be complex. $\endgroup$
    – Nagabhushan S N
    Jul 13, 2020 at 13:32
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    $\begingroup$ @NagabhushanSN $\lambda^2 = 1$ does means $\lambda \pm 1$, but the problem is that $\lambda^2 = 1$ is a false premise. This answer is slightly wrong: $\langle \lambda v, \lambda u \rangle$ is not $\lambda^2 \langle v, v \rangle$ since the first $\lambda$ contributes a $\lambda^\star$. Hence, it should be $|\lambda|^2 = 1$, and thus $\lambda = e^{i \theta}$, not $\lambda = \pm 1$. For example, this is true for $2 \times 2$ rotation matrices. However, the rest of the proof works, with the $- \langle v_1, v_2 \rangle$ replaced by $e^{i(\theta_2 - \theta_1)} \langle v_1, v_2 \rangle$. $\endgroup$
    – xdavidliu
    Nov 25, 2020 at 11:34
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    $\begingroup$ This answer is inaccurate and requires a fix. $\bar{\lambda_1} \lambda_2 = 1$ in case $\bar{\lambda_1} = e^{i\theta}$ and $\lambda_2 = e^{-i\theta}$. Thus, $\lambda_1 = \lambda_2$ and either this is a degenerate case of eigenvalues with multiplicity 2 or more, or $<u,v> = 0$. $\endgroup$
    – Boris Burkov
    Aug 15, 2021 at 1:56
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I believe it is symmetric matrices, not orthogonal ones, which have orthogonal eigenvectors from distinct eigenvalues.

Take any $n$-by-$n$ matrix $A$ and let $u_i$ be an $n$-by-$1$ right eigenvector with eigenvalue $\lambda_i$. Also let $v_j$ be a $1$-by-$n$ left eigenvector with eigenvalue $\lambda_j$. Then: $$Au_i=\lambda_iu_i$$ $$v_jA=\lambda_jv_j$$ With these two equations we can show both of: $$v_jAu_i=\lambda_iv_ju_i$$ $$v_jAu_i=\lambda_jv_ju_i$$ So that, by subtracting the above expressions: $$(\lambda_i-\lambda_j)v_ju_i=0$$ Therefore, if the two eigenvalues are distinct, the left and right eigenvectors must be orthogonal. If $A$ is symmetric, then the left and right eigenvectors are just transposes of each other (so we can think of them as the same). Then the eigenvectors from different eigenspaces of a symmetric matrix are orthogonal.

Edit: The restriction to symmetric matrices is valid for real vector spaces; the situation is more complex with complex vector spaces. :)

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    $\begingroup$ The spectral theorem applies to all normal matrices, and orthogonal matrices are normal. $\endgroup$
    – Ian
    Oct 14, 2015 at 21:42
  • $\begingroup$ @Ian - Thanks for that clarification. I'm not as familiar with complex linear algebra so I was answering from a perspective of real vector spaces. $\endgroup$
    – Matt Dickau
    Oct 14, 2015 at 21:44
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    $\begingroup$ It still works when $Q$ are real and normal. It's just that if a matrix is normal and not Hermitian then at least one eigenvalue is complex. But the eigenvectors for distinct eigenvalues are still orthogonal. $\endgroup$
    – Ian
    Oct 14, 2015 at 21:46
  • $\begingroup$ I wouldn't consider complex eigenvalues to be valid solutions to the eigenvalue problem for vectors from a real vector space - though they are certainly valid solutions for the complexification of that space. $\endgroup$
    – Matt Dickau
    Oct 14, 2015 at 21:56
  • $\begingroup$ This answer is irrelevant. The question was about orthogonal matrices, not symmetric. Eigenvectors of both orthogonal and symmetric matrices are orthogonal, but the question was specifically about orthogonal. $\endgroup$
    – Boris Burkov
    Aug 15, 2021 at 1:50

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