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I was taught that any orthogonal matrix must have eigenvalues of magnitude 1, i.e. $\lambda_i=\pm1,\forall i$. However, I am given the following matrix $$ M = \frac{1}{2}\begin{pmatrix} -1 & -1 & \sqrt{2} \\ 1 & 1 & \sqrt{2} \\ \sqrt{2} & -\sqrt{2} & 0 \end{pmatrix}. $$ We can see it is orthogonal by computing $MM^T$. We also find that $\det M=-1$, so the matrix corresponds to a reflection. Thus we know there is one eigenvalue $\lambda_1=-1$ and the transformation reflects vectors along the plane whose normal is the eigenvector corresponding to $\lambda_1$.

However, trying to find the other two eigenvalues through $\operatorname{tr}{M} = \lambda_1 + \lambda_2 + \lambda_3$ I get $$ \lambda_2 + \lambda_3 = 1, $$ which contradicts $\lambda_i=\pm1$. And indeed when I calculate the characteristic polynomial I get the equation $$ \lambda^3-\lambda+1=0, $$ which is not divisible by $(\lambda+1)$, which is yet another contradiction. But in my textbook it explicitly states that an orthogonal matrix has eigenvalues $\pm1$ and that a reflection has at least one eigenvalue equal to $-1$ (which is just common sense).

Can you help me understand where my reasoning is wrong? And how to solve the remaining eigenvalues and eigenvectors?

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  • $\begingroup$ Tried to edit, but can't fix the formatting of the "\tr" function... $\endgroup$ Commented Feb 19 at 20:08
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    $\begingroup$ Eigenvalues do not have to be real, so $|\lambda| = 1$ does not imply $\lambda = \pm 1$. $\endgroup$ Commented Feb 19 at 20:09
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    $\begingroup$ Recheck your calculation of the characteristic polynomial. (Note that $\lambda=-1$ is not a root of $\lambda^3-\lambda+1=0$, so this can't be the right polynomial.) $\endgroup$ Commented Feb 19 at 20:11
  • $\begingroup$ characteristic polynomial is $x^3 + 1 .$ Indeed $M^3 = -I.$ The determinant is $-1$ but $M^2$ is not the identity, so $M$ does not create a reflection. A null vector for $M + I$ is $ v = (\sqrt 2, 0, -1)^T.$ So $v$ is an eigenvector. What is its eigenvalue? Which is to say, what is $Mv?$ $\endgroup$ Commented Feb 19 at 20:27
  • $\begingroup$ Consider a quarter turn rotation matrix in the $\mathbb{R}^2$. The eigenvalues will be $\pm i$ and $|\pm i|=1$ but the trace will be $0$. $\endgroup$ Commented Feb 20 at 1:13

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