Find the eigenvalues of linear transformation $T(x)= x - 2 \frac{x.v}{v.v} v$
Solution 1:
Your eigenvalues are correct, but your reasoning is still wrong. You can't just define $v$, it is supposed to be arbitrary
ust find the eigenspaces: One is $\mathrm{span}({v})$: Just plug in $v$: $$ T(v) = v - 2v = -v. $$ So $v$ has eigenvalue $-1$.
The other one is $\mathrm{span}(v)^\perp$: Let $x \in \mathbb{R}^3$ such that $x.v = 0$. So $$ T(x) = x. $$ So all vectors in $\mathrm{span}(v)$ have eigenvalue 1. $\mathrm{span}(v)^\perp$ has dimension 2.
Operators like yours are called Householder transformations.