# Eigenvalues and Eigenvectors of The Cross Product Linear Transformation

## Problem 593

We fix a nonzero vector $\mathbf{a}$ in $\R^3$ and define a map $T:\R^3\to \R^3$ by
$T(\mathbf{v})=\mathbf{a}\times \mathbf{v}$ for all $\mathbf{v}\in \R^3$.
Here the right-hand side is the cross product of $\mathbf{a}$ and $\mathbf{v}$.

(a) Prove that $T:\R^3\to \R^3$ is a linear transformation.

(b) Determine the eigenvalues and eigenvectors of $T$.

## Proof.

### (a) Prove that $T:\R^3\to \R^3$ is a linear transformation.

To prove that $T$ is a linear transformation, we need to verify the following equalities for all $\mathbf{u}, \mathbf{v}\in \R^3$ and $c\in \R$:

1. $T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})$
2. $T(c\mathbf{v})=cT(\mathbf{v})$.

These follow from the basic properties of cross products as follows.
We have
\begin{align*}
T(\mathbf{u}+\mathbf{v})&=\mathbf{a}\times (\mathbf{u}+\mathbf{v})\\
&=\mathbf{a}\times \mathbf{u}+\mathbf{a}\times \mathbf{v} &&\text{the cross product is distributive}\\
&=T(\mathbf{u})+T(\mathbf{v}).
\end{align*}
As the cross product is compatible with scalar multiplication, we also have
\begin{align*}
T(c\mathbf{v})&=\mathbf{a}\times (c\mathbf{v})=c(\mathbf{a}\times \mathbf{v}) =cT(\mathbf{v}).
\end{align*}
Therefore $T$ is a linear transformation.

### (b) Determine the eigenvalues and eigenvectors of $T$.

Let $\lambda$ be an eigenvalue of the linear transformation $T$ and let $\mathbf{x}$ be an eigenvector corresponding to $\lambda$.
Then we have
$T(\mathbf{x})=\mathbf{a}\times \mathbf{x}=\lambda \mathbf{x}.$ We take the dot product and obtain
\begin{align*}
(a\times \mathbf{x})\cdot \mathbf{x}=\lambda \mathbf{x}\cdot \mathbf{x}=\lambda \|\mathbf{x}\|^2.
\end{align*}
Note that the vector $\mathbf{a}\times \mathbf{x}$ and $\mathbf{x}$ is orthogonal, hence the left-hand side of the above equality is zero.
On the other hand, since $\mathbf{x}$ is an eigenvector, it is a nonzero vector. Hence the length $\|\mathbf{x}\|\neq 0$.
It follows that we have $0=\lambda \|\mathbf{x}\|^2$ with $\|\mathbf{x}\|\neq 0$, and thus $\lambda=0$.
Hence the eigenvalue of $T$ is $\lambda=0$.

This yields that we have $\mathbf{a}\times \mathbf{x}=0\cdot \mathbf{x}=\mathbf{0}$ for any eigenvector $\mathbf{x}$.
Note that the cross product is zero indicates that the vectors $\mathbf{a}$ and $\mathbf{x}$ are parallel.
Hence $\mathbf{x}=r\mathbf{a}$ for a nonzero $r\in \R$.

In summary, the eigenvalue of $T$ is $0$ and corresponding eigenvectors are $r\mathbf{a}$ for any nonzero real number $r\in \R$.

## Comment.

Let $\{\mathbf{e}_1, \mathbf{e}_2\mathbf{e}_3\}$ be the standard basis for $\R^3$ and write $\mathbf{a}=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}$.
Then we have
\begin{align*}
T(\mathbf{e}_1)&=\begin{bmatrix}
a_1 \\
a_2 \\
a_3
\end{bmatrix}\times \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix}
=\begin{bmatrix}
0 \\
a_3 \\
-a_2
\end{bmatrix}\6pt] T(\mathbf{e}_2)&=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}\times \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} =\begin{bmatrix} -a_3 \\ 0 \\ a_1 \end{bmatrix}\\[6pt] T(\mathbf{e}_3)&=\begin{bmatrix} a_1 \\ a_2 \\ a_3 \end{bmatrix}\times \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} =\begin{bmatrix} a_2 \\ -a_1 \\ 0 \end{bmatrix}. \end{align*} It follows that the matrix representation of the linear transformation T with respect to the standard basis is given by \[A=\begin{bmatrix} 0 & -a_3 & a_2 \\ a_3 &0 &-a_1 \\ -a_2 & a_1 & 0 \end{bmatrix}. Note that the matrix $A$ is a real skew-symmetric matrix.
We know that each eigenvalue of a skew-symmetric matrix is either $0$ or purely imaginary.
(See the post “Eigenvalues of Real Skew-Symmetric Matrix are Zero or Purely Imaginary and the Rank is Even“.)
Also if the degree of the matrix is odd, it has $0$ as an eigenvalue.
(See the post “The Determinant of a Skew-Symmetric Matrix is Zero“.)

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