I was wondering if a non-symmetric orthogonal matrix can have his 3 eigenvalues in the real numbers.
All the 3 real eigenvalues orthogonal matrix i've found are symmetric.
Can someone give me a example?
Can an orthogonal non-symmetric 3×3 matrix have 3 real eigenvalues
eigenvalues-eigenvectorslinear algebraorthogonal matricessymmetric matrices
Best Answer
The eigenvalues of an orthogonal matrix of the shape $3\times 3$ (and one can make things more general, but i am not willing to type bigger matrices in the sequel...) have modulus one, being real as stated means that these are among $-1, +1$.
If all eigenvalues are equal, then $A$ is diagonalized in the form $A=SDS^{-1}$ with $D=\pm 1$ in the center, so $A=\pm 1$, this is symmetric.
Else, suppose the eigenvalues are $1,1,-1$. (In this order.) (Pass to $-A$ in the case $-1,-1,1$.) Fix $u_1,u_2$ orthogonal eigenvectors for $1$, and some $u_3$ for $-1$. We may and do assume that $u_1,u_2,u_3$ all have norm $1$, so they form an orthonormal basis of $\Bbb R^3$. We regard the as column vectors.
Using block matrix computations, let $U=[u_1u_2u_3]$ be the matrix pasted together from the vectors as columns, then $$ AU=A[u_1u_2u_3] =[\ Au_1\ Au_2\ Au_3\ ] =[\ u_1\ u_2\ (-u_3)\ ] =U\begin{bmatrix} 1&&\\&1&\\&&-1\end{bmatrix}\ . $$ So $$ A= U\begin{bmatrix} 1&&\\&1&\\&&-1\end{bmatrix} U' $$ is symmetric.
(Note that $U^{-1}=U'$.)