Let $ v \in V$ be a unit vector in a Euclidean vector space. Prove that the endomorphism $$\phi: V \to V, \qquad \phi(x)=x - 2<x,v> v$$ is a reflection.
I know that a reflection is an orthogonal transformation with determinant = -1. I tried proving that the above mapping preserves lengths by showing that $<\phi(x), \phi(y)> = <x,y>$ but my attempts have failed.I feel like I'm missing something.
A hint would be greatly appreciated.
On
Thanks for the hint. This is what I've got: Since $V=\langle v\rangle\oplus \langle v\rangle^\perp$, we can write any vector $x \in V$ uniquely as some linear combination $ x = \alpha v + \sum_{i=1}^r \beta_i w_i $ where $\langle v\rangle^\perp = \langle w_1,...,w_r \rangle$ for an orthonormal basis $(w_1,...,w_r)$. Then the effect of $\phi$ on x is the following $$ \phi(x) = ... = - \alpha v + \sum_{i=1}^r \beta_i w_i $$ using the fact that $w_i$ and $v$ are orthogonal and that $v$ is a unit vector in the intermediate steps. Then we can see that $\phi(v)=-v$ and $\phi( w_i )= w_i$ for all $i$. From this it follows that the matrix of $\phi$ in the orthonormal basis $(v,w_1,...,w_r)$ of V is given by $diag(-1,1,...,1)$, which is clearly orthogonal with determinant -1.
Is the proof correct?
Hint: Write $V=\langle v\rangle\oplus \langle v\rangle^\perp$.