I'm a little confused with the following situation because inner products show up in this linear transformation.:
Take $\mathbb{R^3}$ with the standard inner product. Let $a,b \in \mathbb{R^3}$ so that $\langle a,b\rangle = 2$
$L: \mathbb{R^3} \rightarrow \mathbb{R^3}:x \longmapsto x – \langle x,a\rangle b$
How do I determine the eigenvalues of $L$? I want to see if $L$ is diagonalizable.
I always use a matrix to determine the eigenvalues but it doesn't seem the best idea to do that in this case.
Thank you in advance.
Best Answer
Let $c\in\mathbb R^3$ such that $c$ is orthogonal to both $a$ and $b$. Then $\{a,b,c\}$ is a basis of $\mathbb R^3$ and the matrix of $L$ with respect to this basis is$$\begin{bmatrix}1&0&0\\\lVert a\rVert^2&-1&0\\0&0&1\end{bmatrix}.$$Can you take it from here?