$v,w\in V$ are nonzero vectors s.t. $(v,w)=0$. Define $T:V\to V$ as: $\forall u\in V$.$T(u)=(u,v)w+(u,w)v.$ Find eigenvalues and eigenvectors of $T$

Question

Problem: Let $V$ denote an inner product vector space over the field $F .$ Let $v, w \in V$ be two nonzero vectors such that $(v, w)=0$. Define the map $T: V \rightarrow V$ as follows. For all $u \in V$ set $T(u)=(u, v) w+(u, w) v .$ Find all the eigenvalues and eigenvectors of $T .$

Attempt:
$ Tv = (v,v)w + (v,w)v = (v,v)w = ||v||^2 w $
$ Tw = (w,v)w + (w,w)v = (w,w)v = ||w||^2 v $
Since $ v,w \neq 0_V $ then $ \frac{Tw}{||w||^2} = v $ , $ \frac{Tv}{||v||^2} = w $
So $ T^2v = ||w||^2 ||v||^2 v $, $ T^2w = ||w||^2 ||v||^2 w $ ( by plugging that last two equations into each other )
Hence $ ||w||\cdot||v|| $ is an eigenvalue of $ T $ with the corresponding eigenvector $v,w $ and $ v=w $ [ Now's the problem : since $ v=w $ then from the assumption that $ (v,w) = 0 $ then $ (v,v) =0 $ and $ v=w=0 $ but this is a contradiction to the given that $v,w \neq 0_V $. ]

Questions:

1. How did I get the contradiction that $v=w = 0_V $? what did I do wrong in my attempted proof?
2. Given the failed attempt, how would one prove the above? I don't have any other ideas.

Thanks in advance!

Answer 1

You cannot take $v=w$ since they are orthogonal and non-zero. You have to find $a$ and $b$ such that $av+bw$ is an eigen vector corresponding to the eigen value $\|v\|\|w\|$. This works out to $\|v\|w+\|w\|v$. Similarly, $\|w\|v-\|v\|w $ is an eigen vector corresponding to eigen value $-\|v\|\|w\|$. Also any vector orthogonal to $v$ and $w$ is an eigen vector corresponding to eigen value $0$. The set of all eigen values is $\{\pm \|v\|\|w\|,0\}$ unless the space is two dimensional, in which case $0$ is not an eigen value.