Why are Eigenvectors of an orthogonal matrix with respect to different eigenvalues orthogonal to one another. 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 $1$?
 A: 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 $A^* A = A A^*$.)
A: 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. :)
A: 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.
