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.

  • $\begingroup$ S maps all vectors in V to a scalar multiple of w. This means that S is a rank 1 matrix, with w as the only non- trivial eigenvector. What is the eigenvalue associated with this eigenvector? $\endgroup$
    – Doug M
    Commented Dec 15, 2020 at 5:41
  • $\begingroup$ The eigenvalue would be $\langle w, v \rangle$, right? I don't think I understand what you mean by 'non-trivial eigenvector.' $\endgroup$
    – user862302
    Commented Dec 15, 2020 at 5:48

1 Answer 1


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

$$ f(x) = x^{n-1}(x - \langle w, v \rangle). $$

  • $\begingroup$ This was extremely helpful. Thank you very much. Could you please explain where the formula for $f(x)$ came from, though? Is this only saying that $0$ and $\lambda$ are the eigenvalues of $S$, and the dimension of the eigenspace with $\lambda = 0$ is $n-1$ because $S$ sent $n-1$ vectors in the eigenbasis to $0$? $\endgroup$
    – user862302
    Commented Dec 15, 2020 at 6:36
  • 1
    $\begingroup$ Yes, this is exactly what this is saying. If you'd like to arrive at it more formally, you can also do this from the definition $f(x) = \det(xI - S_{v,w})$. Since determinant is basis-independent, you can choose the basis $w_1, v_2, \dots, v_n$ in which $S_{v,w}$ is diagonal $S_{v,w} = diag(\lambda, 0, \dots, 0)$ and then you can use the fact that determinant of a diagonal matrix is the product of the entries on the diagonal. You'll get $f(x) = x^{n-1}(x - \lambda)$ as above. $\endgroup$ Commented Dec 15, 2020 at 6:42
  • $\begingroup$ Sorry, but I have another question upon reviewing your proof. There are two things I am having trouble seeing. First, why if $\langle w_1, v_1 \rangle \neq 0$, then $w_1, v_2, \ldots, v_n$ is a basis. And, second, why if $\langle w_1, v_1 \rangle = 0$, then $S(x) = aw$. Would you mind explaining? Thank you again for all your help. $\endgroup$
    – user862302
    Commented Dec 16, 2020 at 11:49
  • $\begingroup$ First: All linear combinations of $v_2, \dots, v_n$ are orthogonal to $v_1$ by linearity of inner product, so when $w_1$ isn't then it cannot be a linear combination of $v_2, \dots, v_n$, so it is linearly independent of them. $\endgroup$ Commented Dec 16, 2020 at 17:42
  • $\begingroup$ Second: $S(x)$ is always a multiple of $w$, so if $\langle w, v\rangle = 0$ then $S(S(x)) = 0$ for every $x$. Updated my answer to clarify these two points, your earlier question and also the case when $v$ or $w$ is zero. Have another look. $\endgroup$ Commented Dec 16, 2020 at 17:44

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