Find the eigenvalues of the linear map with inner product

Question

Consider $\mathbb{R}^{3}$ with the standard inner product. Let $a,b \in \mathbb{R}^{3}$ so that $\langle a,b \rangle = 2$. Define the linear map $L: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3} : x \mapsto x- \langle x,a \rangle b$. Find the eigenvalues of $L$.

How can I solve this problem without using the matrix representation of $L$?

Answer 1

Hint: it is easy to see that $L': x \mapsto \langle x,a \rangle b$ has only one eigenvector with non-zero eigenvalue, $\langle a,b \rangle = 2$. Now try to show that if $\{\lambda_1, \dots, \lambda_n\}$ are eigenvalues of a linear map $L'$, then $\{1-\lambda_1, \dots, 1-\lambda_n\}$ are eigenvalues of $I-L': x \mapsto x - L'(x)$.