Eigenvalues in orthogonal matrices

Let $\lambda$ be $A$ eigenvalue and $Ax=\lambda{x}$.

(1) ${x}^{t}Ax=\lambda{x^tx}>0.$ Because $x^tx>0$, then $\lambda>0$

(2) $|\lambda|^2x^tx=(Ax)^{t}Ax={x}^{t}A^{t}Ax=x^tx.$ So $|\lambda|=1$. Then $\lambda=-1$ or $1$.

Attention: $A$ in (2) may have complex eigenvalue with absolute value 1.

The first part of the problem is well solved above, so I want to emphasize on the second part, which was partially solved.

An orthogonal transformation is either a rotation or a reflection. I will focus on 3D which has lots of practical use. Let us then assume that $A$ is an orthonormal matrix in $\mathbb{R}^3 \times \mathbb{R}^3$.

We find that its eigenvalues are either $1, \text{e}^{\pm i \theta}$ for a rotation or $\pm 1$ for a reflection.

1. For a rotation: We have the following sequence of equalities (since $\det A = 1$)

   \begin{eqnarray*} \det(I -A) &=& \det(A) \det (I-A) \\ &=& \det(A^T) \det(I-A) \\ &=& \det(A^T - I) \\ &=& \det(A - I) \\ &=& -\det(I-A) \quad , \text{since 3 is odd}, \end{eqnarray*} So $\det(I-A)=0$ and $\lambda_1=1$ is an eigenvalue of $A$.

   Now, let $u_1$ the unit eigenvector of $\lambda_1$, so $A u_1 = u_1$. We show that the matrix $A$ is a rotation of an angle $\theta$ around this axis $u_1$. Let us form a new coordinate system using $u_1, u_2, u_1 \times u_2$, where $u_2$ is a vector orthogonal to $u_1$, so the new system is right handed (has determinant = 1). The transformation between the old system $\{e_1, e_2, e_3\}$ and the new system is given by a matrix $B$ with column vectors $u_1, u_2$, and $u_3$. So we have that $B e_i = u_i$. Let us call the coordinate transformation (similarity) matrix $C = B^{-1}AB$. Then \begin{eqnarray*} C e_1 = B^{-1} A B e_1 = B^{-1} A u_1 = B^{-1} u_1 = e_1. \end{eqnarray*} The fact that $C e_1 = e_1$ means two things:

   • The first column of $C$ is $(1,0,0)^T$.

   • The first raw of $C$ is $(1,0,0)$.

   Then $C$ is a matrix of the type \begin{eqnarray*} C = \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & a & b \\ 0 & c & d \\ \end{array} \right ) \end{eqnarray*} Since $A$ is orthogonal $C$ is orthogonal and so the vectors $(a,c)^T$ and $(b,d)^T$ are orthogonal and since \begin{eqnarray*} 1 = \theta A = \det C = ad - bc \end{eqnarray*} we have that the minor matrix with entries $a,b,c,d$ is a rotation (orthogonal with determinat 1). Rotations in 2D are of the form \begin{eqnarray*} \left ( \begin{array}{cc} a & b \\ c & d \end{array} \right ) = \left ( \begin{array}{cc} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array} \right ) \end{eqnarray*} then \begin{eqnarray*} B^{-1}A B = \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos \theta & -\sin \theta \\ 0 & \sin \theta & \cos \theta \end{array} \right ) \end{eqnarray*} But since the eigenvalues of $B^{-1}AB$ are the same as those of $A$ we find the eigenvalues of this matrix and those would be the eigenvalues of $A$.

   For the eigenvalues of this matrix we have that

   \begin{eqnarray*} \det ( C - \lambda I)= 0 \implies (\cos \theta - \lambda)^2 + \sin^2 \theta = 0. \end{eqnarray*} That is,

   \begin{eqnarray*} 1 -2 \lambda \cos \theta + \lambda^2 = 0 \end{eqnarray*} Then

   \begin{eqnarray*} \lambda = \frac{2 \cos \theta \pm \sqrt{4 \cos^2 \theta - 4}}{2} = \cos \theta \pm i \sin \theta = \text{e}^{\pm i \theta}. \end{eqnarray*}

This is the interesting result by Euler where he claimed that all rigid body rotations can be written as rotation about one vector (here $u_1$ by some angle $\theta$).

2. Eigenvalues of a reflection. If we repeat all the steps above for a reflection we find that now

   \begin{eqnarray*} B^{-1}A B = \left ( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & \cos 2 \theta & \sin 2 \theta \\ 0 & \sin 2 \theta & -\cos 2 \theta \end{array} \right ) \end{eqnarray*}

   Now
   \begin{eqnarray*} \det ( C - \lambda I)= 0 \implies -(\cos 2 \theta - \lambda)(\cos 2 \theta + \lambda ) - \sin^2 2 \theta = 0. \end{eqnarray*} and then

   \begin{eqnarray*} -\sin^2 2 \theta - \cos^2 2 \theta + \lambda^2 =0 \end{eqnarray*} and from here

   \begin{eqnarray*} \lambda = \pm 1. \end{eqnarray*} .

Please see the following reference which I used for this answer: Orthogonal Matrices

(1):

Let $Av=\lambda v$ with $v\not=0$, i.e. $\lambda$ is an eigenvalue of $A$. Then $$0<v^t Av=\lambda v^t v=\lambda \|v\|^2.$$

Since $\|v\|^2>0$, we get $\lambda>0$.

(2):

You certainly mean that the determinant of $A$ is $\pm 1$, since the statement about the eigenvalues is not true, for consider the orthogonal matrix

$$\begin{pmatrix}\cos\theta & -\sin\theta\\ \sin\theta & \cos\theta\end{pmatrix}$$

This represents a rotation and has therefore complex eigenvalues.

But if $A$ is orthogonal, then $A^tA=AA^t=I$, therefore applying the $\det$ to both sides and using the multiplication law for determinants, we obtain

$$(\det A)^2 = 1$$

Therefore $\det A=\pm 1$.