Find eigenvalues and eigenvectors of a linear map

Question

Let $\underline w$ and $\underline t$ two fixed vectors and $$ T:E_3\longrightarrow E_3, \qquad T(\underline x):=(\underline x\cdot \underline t)\underline w. $$ Find possible eigenvalues and eigenvectors of $T$ without solving any secular equation.

My attempt. By definition, $\lambda$ is an eigenvalue of $T$ if there exists $\underline x\neq0$ such that $$ T(\underline x)=(\underline x\cdot\underline t)\underline w=\lambda\underline x, $$ that is $(\underline x\cdot\underline t)\underline w-\lambda\underline x=\underline 0$. But how can I proceed now?

Thank You

Answer 1

Suppose that $T$ is not trivial, remark that $w$ and $x$ must be colinear, s $w$ is the unique eigenvector and the eigenvalue is $(w.t)$.

Answer 2

Hint If $\bf w$ and both coefficients $({\bf x} \cdot {\bf t})$ and $\lambda$ are nonzero, then the identity $({\bf x} \cdot {\bf t}) {\bf w} - \lambda {\bf x} = {\bf 0}$ says that $\bf w$ and $\bf x$ are linearly dependent.

Answer 3

Assume that $\underline w$ and $\underline t$ are both nonzero, otherwise the problem is trivial. $T$ maps every input to a scalar multiple of $\underline w$, so the only possibilities for eigenspaces of $T$ are the span of $\underline w$ and the kernel of $T$.

For the first, you have by linearity of the dot product $T(k\underline w)=(k\underline w\cdot\underline t)\underline w = (\underline w\cdot\underline t)(k\underline w)$, which gives you one eigenvalue of $T$ and its corresponding eigenspace.

For the second, since $\underline w\ne 0$, $(\underline x\cdot\underline t)\underline w=0$ iff $\underline x\cdot\underline t = 0$, so the kernel of $T$ consists of all vectors orthogonal to $\underline t$. These are all eigenvectors of $T$ with eigenvalue zero.