Eigenvalues of a particular map $\Bbb C^3 \to \Bbb C^3$ involving an inner product

Question

Let $f:\mathbb{C}^3\rightarrow\mathbb{C}^3$ be the linear map $$f(x)=x-i\langle x,v \rangle v ,$$ where $v\in\mathbb{C}^3$ satisfies $\langle v,v \rangle =1$. What are the eigenvalues and eigenvectors of $f$?

The only way I could think of handling this question would be to make a matrix that belongs to $f$ and then calculate eigenvalues/-vectors in the usual way. However this gives us a very big (and frankly ugly) matrix to work with, so I was wondering if there is an easier way to answer this question.

Answer 1

You can choose whatever basis you like!

Let $\{v=v_1,v_2,v_3\}$ be an orthogonal basis for $\mathbb{C}^3$; then \begin{align} f(v_1)&=v-i\langle v,v\rangle v=(1-i)v\\ f(v_2)&=v_2-i\langle v_2,v\rangle v=v_2\\ f(v_3)&=v_3-i\langle v_3,v\rangle v=v_3 \end{align} Thus the eigenvalues are $1-i$ (single) and $1$ (double).

Answer 2

Hint Suppose $x$ is an eigenvector of $f$, say with eigenvalue $\lambda$, so that $$\lambda x = x - i \langle x, v \rangle v .$$ Rearranging gives $$(\lambda - 1) x = - i\langle x , v \rangle v .$$ Hence, if $\lambda \neq 1$, $x$ must be a multiple of $v$.

Additional hint So, if $\lambda \neq 1$, then by rescaling we may assume $x = v$. Then, we have $f(x) = v - i \langle v, v \rangle v = (1 - i) v$, and hence $v$ is an eigenvector of eigenvalue $1 - i$. Conversely, if $\lambda = 1$, then we have $-i \langle x, v \rangle v = 0$, which (since $v \neq 0$) holds iff $\langle x, v \rangle = 0$, so $\langle v \rangle^{\perp}$ is a ($2$-dimensional) eigenspace of $f$ of eigenvalue $1$.