Computing the characteristic polynomial of a linear transformation I am confused as to how to solve this problem, as the linear transformation includes an arbitrary inner product.

Let $V$ be a finite-dimensional inner product space. Fix vectors $v,w \in V$ and define $S_{v,w} (x) = \langle x,v \rangle w$. What is the characteristic polynomial of $S$?

I cannot figure out how to get started, nor do I have any intuition for what the eigenvalues of $S$ are. The vector space is arbitrary, so I cannot choose a basis. With a bit of help, I am in the process of proving that $x$ is an eigenvector of $S$ if and only if $w = ax$ for some scalar $a$ or $\langle x, v \rangle = 0$. In the latter case, $x$ has eigenvalue $0$. In the former case, $x$ has eigenvalue $\langle x, v \rangle a$. Where I am stuck, however, is the following: prove the reverse direction (that these are the only eigenvalues), finding the dimensions of the eigenspaces, and finding exactly how many eigenvectors (or eigenspaces) there are. The factors of the characteristic polynomial are $(z - \lambda_i)^{\dim V_{\lambda_i}}$ where $\dim V_{\lambda_i}$ is the dimension of the eigenspace.
 A: If $v$ or $w$ is the zero vector, then $S_{v,w}$ is the zero map whose characteristic polynomial is $x^n$.
Assume that both $v$ and $w$ are non-zero. Let $w_1 = \frac{w}{\|w\|}$ and complete $w_1$ to an orthonormal basis $w_1, w_2, \dots, w_n$. Let $v_1 = \frac{v}{\|v\|}$ and complete $v_1$ to an orthonormal basis $v_1, v_2, \dots, v_n$.
First, assume that $\langle w_1, v_1\rangle \ne 0$. Since every linear combination of $v_2, \dots, v_n$ is orthogonal to $v_1$, we see that $w_1$ is linearly independent from $v_2, \dots, v_n$. Thus, $w_1, v_2, \dots, v_n$ is a basis. Moreover,
$$
S_{v,w}(w_1) = \lambda w_1 \\
S_{v,w}(v_2) = 0 \\
\dots \\
S_{v,w}(v_n) = 0 \\
$$
where $\lambda = \langle w, v \rangle$. Thus, we have found the basis of eigenvectors of $S_{v,w}$ (and $w$ is indeed an eigenvector as you expect). In this basis, the matrix of $S_{v,w}$ is diagonal with $(\lambda, 0, \dots, 0)$ on the diagonal so its characteristic polynomial is
$$
f(x) = x^{n-1}(x - \lambda).
$$
Now, assume that $\langle w_1, v_1\rangle = 0$. Note that for every $x \in V$, $S_{v,w}(x)$ is a scalar multiple of $w_1$. Therefore, $S_{v,w}(S_{v,w}(x)) = 0$ for every $x \in V$. In other words, $S_{v,w}$ is nilpotent and so its characteristic polynomial is
$$
f(x) = x^n.
$$
Finally, note that these two cases can be written together as
$$
f(x) = x^{n-1}(x - \langle w, v \rangle).
$$
