# Can an orthogonal non-symmetric 3x3 matrix have 3 real eigenvalues?

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?

• Do you meant $3$ distinct real eigenvalues? That's impossible. – José Carlos Santos May 17 '19 at 23:08
• An orthogonal matrix will have real eigenvalues if and only if it is symmetric – Ben Grossmann May 17 '19 at 23:33
• No, i meant if no complex eigenvalues could be found in an orthogonal 3x3 non symmetric matrix – 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 $$U^{-1}=U'$$.)

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