Eigenvalues/vectors for non-specific linear operator? I need to find the eigenvalues and eigenvectors of $A$ in $\mathbb{R}^3$ such that $Ax = |a|^2x-(a \cdot x)a$ where $a$ is a given constant vector.
I found $A-\lambda I$ in the usual $\mathbb{R}^3$ basis: 
\begin{bmatrix}
|a|^2-a_x^2-\lambda & -a_xa_y & -a_xa_z \\
-a_xa_y & |a|^2-a_y^2-\lambda & -a_ya_z \\
-a_xa_z & -a_ya_z & |a|^2-a_z^2-\lambda
\end{bmatrix}
The determinant of this matrix is $(\lambda)(\lambda-x^2-y^2-z^2)^2$ (Wolfram). This gives me eigenvalues of $0$ and $|a|^2$. Is this correct? If so, how do I find the eigenvectors in this system? Typically for the $\lambda=0$ case I would simply reduce A to row-echelon form and find the vector. Setting $\lambda=0$ gives me:
\begin{bmatrix}
a_y^2 + a_z^2 & -a_xa_y & -a_xa_z \\
-a_xa_y & a_x^2 + a_z^2 & -a_ya_z \\
-a_xa_z & -a_ya_z & a_x^2 + a_y^2
\end{bmatrix}
I don't see how to reduce this matrix or proceed from here to find eigenvectors.
 A: I think it's easier not to use matrix representation.
Suppose $x$ is an eigenvector. Then $Ax = \lambda x$, i.e.,
$$|a|^2 x - (a.x)a = \lambda x \ \ \hbox{ or } \ \ (|a|^2 - \lambda)x = (a.x) a$$
If $x$ is parallel with $a$, then $x = a$ is an eigenvector and 
$$(|a|^2 - \lambda)a = |a|^2a \ \ \Longrightarrow  \ \ \lambda = 0.$$
That is, $a$ is a basis vector for the kernel of $A$. Inspecting $A$, it is the only basis vector.
If $x$ and $a$ are not parallel, it must be that $a.x = 0$ with eigenvalue $\lambda = a^2$. I'll leave it to you to find the corresponding eigenvectors/eigenspace.
A: We can write
$$
\mathsf{A}\mathbf{\cdot x}=\mathbf{|a|}^{2}\{\mathsf{U}-\mathbf{e}_{3}
\mathbf{e}_{3}\}\cdot \mathbf{x}
$$
where $\mathsf{U}$ is the unit $3\times 3$-matrix and $\mathbf{e}_{3}$ the
unit vector along $\mathbf{a}$. With $\{\mathbf{e}_{1},\mathbf{e}_{2},
\mathbf{e}_{3}\}$ an orthonormal basis, we have
$$
\mathsf{U}=\sum_{j=1}^{3}\mathbf{e}_{j}\mathbf{e}_{j}=\mathsf{U}_{1,2}+
\mathbf{e}_{3}\mathbf{e}_{3}
$$
so $\mathsf{A}$ decomposes in the direct sum of $\mathbf{|a|}^{2}$ times the
unit operator on the subspace spanned by $\mathbf{e}_{1}$ and $\mathbf{e}_{2}
$ and the zero operator on the subspace spanned by $\mathbf{e}_{3}$. Thus
there are two independent eigenvectors   $\mathbf{e}_{1}$ and $\mathbf{e}_{2}
$ (or appropriate linear combinations) at the eigenvalue $\mathbf{|a|}^{2}$.
By the way, another form of $\mathsf{A}\mathbf{\cdot x}$ is  $\mathbf{a}%
\times (\mathbf{a\times x)}$.
