# Eigenvalues of Orthogonal Matrices Have Length 1. Every $3\times 3$ Orthogonal Matrix Has 1 as an Eigenvalue

## Problem 419

**(a)** Let $A$ be a real orthogonal $n\times n$ matrix. Prove that the length (magnitude) of each eigenvalue of $A$ is $1$.

**(b)** Let $A$ be a real orthogonal $3\times 3$ matrix and suppose that the determinant of $A$ is $1$. Then prove that $A$ has $1$ as an eigenvalue.

Sponsored Links

Contents

## Proof.

### (a) Prove that the length (magnitude) of each eigenvalue of $A$ is $1$

Let $A$ be a real orthogonal $n\times n$ matrix.

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{v}$ be a corresponding eigenvector.

Then we have

\[A\mathbf{v}=\lambda \mathbf{v}.\]
It follows from this we have

\[\|A\mathbf{v}\|^2=\|\lambda \mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2.\]
The left hand side becomes

\begin{align*}

&\|A\mathbf{v}\|^2\\

&=\overline{(A\mathbf{v})}^{\trans}(A\mathbf{v}) && \text{by definition of the length}\\

&=\bar{\mathbf{v}}^{\trans}A^{\trans}A\mathbf{v} && \text{because $A$ is real}\\

&=\bar{\mathbf{v}}^{\trans}\mathbf{v} && \text{because $A^{\trans}A=I$ as $A$ is orthogonal}\\

&=\|\mathbf{v}\|^2 && \text{by definition of the length.}

\end{align*}

It follows that we obtain

\[\|\mathbf{v}\|^2=|\lambda|^2\|\mathbf{v}\|^2.\]
Since $\mathbf{v}$ is an eigenvector, it is non-zero, and hence $\|\mathbf{v}\|\neq 0$.

Canceling $\|\mathbf{v}\|$, we have

\[|\lambda|^2=1.\]
Since the length is non-negative, we obtain

\[|\lambda|=1,\]
as required.

### (b) Prove that $A$ has $1$ as an eigenvalue.

Let $A$ be a real orthogonal $3\times 3$ matrix with $\det(A)=1$.

Let us consider the characteristic polynomial $p(t)=\det(A-tI)$ of $A$.

The roots of $p(t)$ are eigenvalues of $A$.

Since $A$ is a real $3\times 3$ matrix, the degree of the polynomial $p(t)$ is $3$ and the coefficients are real.

Thus, there are two cases to consider:

- there are three real eigenvalues $\alpha, \beta, \gamma$, and
- there is one real eigenvalue $\alpha$ and a complex conjugate pair $\beta, \bar{\beta}$ of eigenvalues.

Let us first deal with case 1.

By part (a), the lengths of eigenvalues $\alpha, \beta, \gamma$ are $1$. Since they are real numbers, we have

\[\alpha=\pm 1, \beta=\pm 1, \gamma=\pm 1.\]
Recall that the product of all eigenvalues of $A$ is the determinant of $A$.

(For a proof, see the post “Determinant/trace and eigenvalues of a matrix“.)

Thus we have

\[\alpha \beta \gamma=\det(A)=1.\]
Thus, at least one of $\alpha, \beta, \gamma$ is $1$.

Next, we consider case 2. Again the lengths of eigenvalues $\alpha, \beta, \bar{\beta}$ are $1$.

Then we have

\begin{align*}

1&=\det(A)=\alpha \beta \bar{\beta}\\

&=\alpha |\beta|^2=\alpha.

\end{align*}

Therefore, in either case, we see that $A$ has $1$ as an eigenvalue.

Add to solve later

Sponsored Links