Let $V$ a vector space with $\dim V < \infty$, and $U\subset V$. let $T_U:V\rightarrow V$ with $T_U(v)=2s-v$ where $s$ is the orthogonal projection of $v$ onto $U$. I want to prove that $\forall v_1,v_2\in V \ :\ \langle T_U(v_1) \ | \ T_U(v_2)\rangle = \langle v_1 \ | \ v_2\rangle$.

I tried to do the following:

Let $v_1, v_2 \in V$, and let $s_1$ to be the orthogonal projection of $v_1$ onto $U$ and $s_2$ the orthogonal projection of $s_2$ onto $U$. then:

$\langle T_U(v_1) \ | \ T_U(v_2)\rangle = \langle 2s_1-v_1 \ | \ 2s_2 - v_2\rangle = \langle 2s_1 \ | \ 2s_2\rangle - \langle 2s_1 \ | \ v_2\rangle - \langle v_1 \ | \ 2s_2\rangle + \langle v_1 \ | \ v_2\rangle$

but from there I don't know how to continue..


The first three terms are rewritten $$2\langle s_1|s_2-v_2\rangle+2\langle s_1-v_1|s_2\rangle$$ Can you do the rest of the job? Show that each term is equal to zero!


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