Find the adjoint of an operator

Question

Let $V$ be an inner product space over $\mathbb{R}$, and $0\neq v \in V$ some vector.

$T: V \rightarrow V$ is the operator defined as: $$ T(x) = x- \frac{2\left \langle v,x\right \rangle }{\left \langle v,v \right \rangle }v $$

Find the adjoint of $T$, $T^{*}$.

I tried using the definition of the adjoint, by taking some vectors $w,q$, place in $\left \langle Tw,q \right \rangle$, and getting to a form $\left \langle w,T^{*}q \right \rangle$, where $T^{*}$ is written in explicit form, but i got stuck along the way.

Would appreciate any help.

Answer 1

We have

$$\langle Tx,y\rangle = \left\langle x - \frac{2\langle v,x\rangle}{\langle v,v\rangle}v, y\right\rangle = \langle x,y\rangle - 2\frac{\langle v,x\rangle\langle v,y\rangle}{\langle v,v\rangle} = \left\langle x,y - \frac{2\langle v,y\rangle}{\langle v,v\rangle}v\right\rangle = \langle x,Ty\rangle$$

therefore $T^* = T$.

Answer 2

To simplify, we can assume $\|v\|^2=\langle v,v\rangle =1$, just take $v_1:=\frac v{\|v\|}$, then $$T(x)=x-2\langle v_1,x\rangle v_1$$ Note that, as $v_1\ \perp\ (x-\langle v_1,x\rangle v_1)\,=:y$, geometrically $T(x)$ will be just the reflection of $x$ along the hyperplane ${v_1}^\perp$, because $y\in {v_1}^\perp\ $ and $\ y+\alpha v_1=x\ $ and $\ y-\alpha v_1=T(x)\ $ with the same $\alpha=\langle v_1,x\rangle$.

It suggests that $T$ will be self-adjoint.

Anyway, we can just start out from the definition: $$\langle x,\,T(y)\rangle =\big\langle x,\ y-2\langle v_1,y\rangle v_1\big\rangle=\langle x,y\rangle -2\langle v_1,y\rangle\langle x,v_1\rangle$$ and we want to write it of the form $\langle a,y\rangle$ for some $a$.
If such an equation is satisfied by all vectors $y$, then we must have $T^*(x)=a$.

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Another way to approach the problem is to reduce it to easier pieces, by applying $(A+\lambda B)^*=A^*+\lambda B^*$ with $A=\mathrm{Id}$ and $B=x\mapsto \langle v,x\rangle v$. Then, basically you only have to calculate $B^*$.

Answer 3

$Px = \frac{\langle v,x\rangle}{\|v\|^2}v$ is the orthogonal projection of $x$ onto the unit vector $\frac{1}{\|v\|}v$. So $P^*=P=P^2$. And $T=I-2P$, which makes it selfadjoint, i.e., $T^*=T$.