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?

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    $\begingroup$ Do you meant $3$ distinct real eigenvalues? That's impossible. $\endgroup$ – José Carlos Santos May 17 '19 at 23:08
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    $\begingroup$ An orthogonal matrix will have real eigenvalues if and only if it is symmetric $\endgroup$ – Ben Grossmann May 17 '19 at 23:33
  • $\begingroup$ No, i meant if no complex eigenvalues could be found in an orthogonal 3x3 non symmetric matrix $\endgroup$ – Marco Villalobos May 18 '19 at 10:29

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 $U^{-1}=U'$.)

  • $\begingroup$ You're assuming that if $A$ is orthogonal then $A$ is diagonalizable, which requires some form of the spectral theorem $\endgroup$ – Ben Grossmann May 17 '19 at 23:40
  • $\begingroup$ Another approach, once it's established that orthogonal matrices are diagonalizable, is to note that $A$ is diagonalizable with eigenvalues in $\{-1,1\}$ if and only if $A^2 = I$. $\endgroup$ – Ben Grossmann May 17 '19 at 23:42
  • $\begingroup$ I was trying to put some details, so that the argumentation works without heavy metal stuff. Where i'm assuming the diagonalizability? $\endgroup$ – dan_fulea May 17 '19 at 23:42
  • $\begingroup$ "Fix $u_1,u_2$ orthogonal eigenvectors for $1$, and some $u_3$ for $-1$" $\endgroup$ – Ben Grossmann May 17 '19 at 23:43
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    $\begingroup$ We have $A=UDU'$ with a diagonal, thus symmetric $D$. Then $A'=(UDU')'=(U')'D'U'=UDU'=A$. $\endgroup$ – dan_fulea May 18 '19 at 15:24

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