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. Add to solve later

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:

1. there are three real eigenvalues $\alpha, \beta, \gamma$, and
2. 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

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