Question

We have seen that an orthogonal matrix with determinant $\mathbf{1}$ has at least one eigenvalue $\mathbf{1}$, and an orthogonal matrix with determinant $\mathbf{-}\mathbf{1}$has at least one eigenvalue . Show that the other two eigenvalues in both cases are ${\mathit{e}}^{\mathbf{i}\mathbf{\theta }}\mathbf{,}{\mathit{e}}^{\mathbf{-}\mathbf{i}\mathbf{\theta }}$, which, of course, includes the real values $\mathbf{1}$ (when $\mathbf{0}\mathbf{=}\mathbf{0}$), and $\mathbf{-}\mathbf{1}$ (when $\mathbf{\theta }\mathbf{=}\mathit{\Pi }$). Hint: See Problem 9, and remember that rotations and reflections do not change the length of vectors so eigenvalues must have absolute value $\mathbf{1}$.

Short answer

Eigen values are $e^{+i\theta }$

Step-by-step solution

Given Information

The eigenvalues of $3\times 3$orthogonal matrices $M$with determinant $\pm 1$. Say that the matrix M can be diagonalized as

$C^{-1}MC=D$

Where, $D$is diagonal. If we take the determinant of the former relation.

Orthogonal Matrix

A square matrix with real numbers or elements is said to be an orthogonal matrix if its transpose is equal to its inverse matrix. Or when the product of a square matrix and its transpose gives an identity matrix, then the square matrix is known as an orthogonal matrix.

Suppose $A$ is a square matrix with real elements and of $nxn$ order and $A$ is the transpose of $A$. Then according to the definition, if, $A=A^1$ is satisfied, then,

$A{A}^T=1$

Where ' $\mathit{I}$' is the identity matrix, ${\mathit{A}}^{\mathbf{-}\mathbf{1}}$is the inverse of matrix $\mathit{A}$, and ' role="math" localid="1664196263816" $\mathit{n}$' denotes the number of rows and columns.

Orthogonal Matrix

In this problem we find the eigenvalues of $3\times 3$orthogonal matrices $M$ with determinant $\pm 1$. Say that the matrix $M$can be diagonalized as

$C^{-1}MC=D$

Where, $D$ is diagonal. If we take the determinant of the former relation, we have

$$\begin{gathered} \det \left(C^{-1}MC\right)=\det C^{-1}\det M\det C \\ =\frac{1}{\det C}\det M\det C \\ =\det M \\ =\det D, \end{gathered}$$

Where, we have used the property that the determinant of a product of matrices is the product of determinants, and that the determinant of the inverse of a matrix is the inverse of the determinant. Since the matrix $D$is diagonal, its determinant is the product of all eigenvalues. We therefore have

${\lambda }_1{\lambda }_2{\lambda }_3=\pm 1$

Since, $\det M=\pm 1$. We know that for an orthogonal matrix with determinant $\pm 1,$we have one eigenvalue which is $\pm 1$, respectively. Say that this is ${\lambda }_1$. We therefore have

${\lambda }_2{\lambda }_3=1$in both cases. Since orthogonal matrices preserve the length of a vector ${\mathit{r}}^{\mathbf{T}}\mathit{r}$, we can write

$(M\mathbf{r}{)}^{\text{T}}\left(M\mathbf{r}\right)={\lambda }^{*}\lambda {\mathbf{r}}^{\text{T}}\mathbf{r}={\mathbf{r}}^{\text{T}}\mathbf{r}$

Which gives $|\lambda {|}^2=1$. Since the magnitude is non-negative, $\left|\lambda \right|=1$. Given that the magnitude of both eigenvalues is one, they should have the form

${\lambda }_k=e^{i{\theta }_k}.$

Since their product is one, we have

$$\begin{gathered} e^{i{\theta }_1}{e}^{i{\theta }_2}=1 \\ {\theta }_1+{\theta }_2=0 \end{gathered}$$

We therefore have the eigenvalues

$e^{\pm i\theta },$

Where, we have renamed ${\theta }_1=-{\theta }_2=\theta$.