To prove that $A$ has a one-dimensional eigenspace , where $A \in SO(3)$ , $A \ne I$

Question

Let $A\ne I$ be a $3\times3$ real orthogonal matrix with determinant $1$ , then how to prove that $A$ has a one-dimensional eigenspace ?

Answer 1

I would prove it like this:

$A$ can be construed as acting on $\Bbb C^3$ in the usual manner: if we write $z \in \Bbb C^3$ as $z = x + iy$, where $x, y \in \Bbb R^3$, then since $A$ is real we have $Az = Ax + iAy$; this is a convenient device in the present contest since it allow us to discuss the eigenvalues of $A$; I claim they are all unimodular complex numbers. For if $Az = \lambda z$ with $0 \ne z \in \Bbb C^3$, then since $A^TA =I$ carries over to $A^\dagger A = I$ by virtue of the fact that $A$ is real, we have

$\langle z, z \rangle = \langle Az, Az \rangle = \langle \lambda z, \lambda z \rangle = \langle z,\bar \lambda \lambda z \rangle = \bar \lambda \lambda \langle z, z \rangle; \tag{1}$

since $\langle z, z \rangle \ne 0$ we have $\bar \lambda \lambda = 1$. Now since $A$ is real and its characteristic polynomial is of degree 3, $A$ must have at least one real eigenvalue $\lambda_1$; if $\lambda_2$, $\lambda_3$ are the other eigenvalues, then $\lambda_1 \lambda_2 \lambda_3 = \det(A) = 1$. Now either $\lambda_2$, $\lambda_3$ are either both real, or they form a complex conjugate pair. If real, then the fact they are unimodular (so they can only take the values $\pm 1$) coupled with $\lambda_1 \lambda_2 \lambda_3 = 1$ forces at least two of the three eigenvalues to be of the same sign. If two are positive, then the third must also be positive since their product is $1$; then $\lambda_i = 1$, $1 \le i \le 3$; this case is ruled out since it corresponds to the forbidden case $A = I$; thus at least two must be negative; but then the third must be $+1$ to preserve the value of the deteminant at $+1$. The eigenspace corresponding to $1$ is thus one-dimensional. In the event that $\lambda_2$, $\lambda_3$ form a complex conjugate pair, we have $\lambda_2 \lambda_3 = 1$ which forces $\lambda_1 = 1$ and so there is a one-dimensional eigenspace in this case as well.

Answer 2

The characteristic polynomial $P_A(x)$ has degree $3$ and real coefficients and hence has at least one real root, that is, at least one real eigenvalue. Consider a nonzero eigenvector corresponding to that eigenvalue.

Works for odd instead of $3$.

Answer 3

We have $A^T=A^{-1}$. Therefore $$ \begin{aligned} \det(A-I)&=\det(A-AA^T)=\det(A(I-A^T))\\ &=\det A\cdot\det(I-A^T)\\ &=\det(I-A^T)=\det((I-A^T)^T)\\ &=\det(I-A)=-\det(A-I), \end{aligned} $$ where the last equality uses the fact that as $3$ is an odd number all the $3\times3$ matrices $X$ satisfy $\det(-X)=(-1)^3\det X=-\det X.$

This proves that $\lambda=1$ is an eigenvalue of $A$.

The corresponding eigenspace cannot be 3-dimensional, because then we would have $A=I$. It cannot be 2-dimensional either. For that eigenspace would be a plane $T$ through the origin. Because $A$ is orthogonal, the normal $N$ (= the orthogonal complement, $N=T^\perp$) of $T$ would also be stable under $A$, and hence (being 1-dimensioinal) an eigenspace of $A$. The eigenvalue of $A$ on $N$ would then also have to be $=1$, because the product of the eigenvalues $=\det A=1$. Thus the 2-dimensional eigenspace would actually be 3-dimensional.