Q:

Find Eigenvalues and Eigenvectors of A

Let $\mathbf{A}\mathbf{x}=\mathbf{a} \times \mathbf{x}$, where $\mathbf{x} $ and $\mathbf{a}$ are in R$^3$ and $\mathbf{a}$ is a fixed or constant vector. Find the eigenvalues and eigenvectors of A. 

A:

I'm going to assume that you meant $A\vec{x} = \vec{a}\times\vec{x}$ (not $A\vec{b} = \vec{a}\times\vec{x}$). Here's a hint: what do you know about $\vec{a}\times\vec{x}$ with regards to $\vec{x}$ (geometrically)?

(I'm going to assume that $\vec{a}\neq \vec{0}$ since otherwise $A$ is the zero matrix which is kind of silly for the purposes of eigenvalues/eigenvectors.) We can view $\vec{a}\times\vec{x}$ as the matrix product
$$\left(\begin{array}{rrr} 0 & -a_3 & a_2 \\ a_3 & 0 & -a_1 \\ -a_2 & a_1 & 0 \end{array}\right)\left(\begin{array}{c} x_1 \\ x_2 \\ x_3\end{array}\right).$$
You can see this since $\vec{a}\times\vec{x} = (a_2 x_3 - a_3 x_2, a_3 x_1 - a_1 x_3, a_1 x_2 - a_2 x_1)$ and the matrix product above works out exactly the same. Using the usual approach, we want to solve
$$A\vec{x}= \lambda\vec{x}.$$
i.e. we want to solve
$$\left(\begin{array}{rrr} -\lambda & -a_3 & a_2 \\ a_3 & -\lambda & -a_1 \\ -a_2 & a_1 & -\lambda \end{array}\right)\left(\begin{array}{c} x_1 \\ x_2 \\ x_3\end{array}\right) = \left(\begin{array}{c} 0 \\ 0 \\ 0\end{array}\right).$$
Naturally we want $\det(A-\lambda I) = 0$ so that we can have eigenvalues and eigenvectors. If we expand the determinant of the above what we get is
$$0 = -\lambda\left|\begin{array}{rr} -\lambda & -a_1 \\ a_1 & -\lambda\end{array}\right| + a_3\left|\begin{array}{rr} a_3 & -a_1 \\ -a_2 & -\lambda\end{array}\right| + a_2\left|\begin{array}{rr} a_3 & -\lambda \\ -a_2 & a_1\end{array}\right|.$$
Working this out further we have
$$0 =-\lambda(\lambda^2 + a_1^2) - a_3(a_3\lambda+a_1a_2) + a_2(a_1a_3-a_2\lambda)$$
which becomes
$$0 = -\lambda^3 -a_1^2\lambda - a_3^2\lambda - a_1a_2a_3 + a_1a_2a_3 -a_2^2\lambda = -\lambda^3 - (a_1^2+a_2^2+a_3^2)\lambda.$$
Doing some factoring we get
$$0 = \lambda(\lambda^2 + a_1^2+a_2^2+a_3^2).$$
Since $\vec{a}\neq \vec{0}$, $a_1^2+a_2^2+a_3^2 > 0$ so we cannot have that $\lambda^2 = -a_1^2-a_2^2-a_3^2$. Thus the only choice is $\lambda = 0$. I'll leave it to you to figure out what the eigenvectors are.

