Find the adjoint operator $T^*$ of $T:V\to V$ defined by $Tx=x-\frac{2<v,x>}{||v||^2}v$ for some fixed $v\in V$, where $V$ is under $\mathbb{R}$.

Question

Find the adjoint operator $T^*$ of $T:V\to V$ defined by $Tx=x-\frac{2<v,x>}{||v||^2}v$ for some fixed $v\in V$, where $V$ is under $\mathbb{R}$.

Hello all. I would like to find $T^*$ such that $<T\alpha, \beta>=<\alpha,T^*b> $ for all $\alpha, \beta\in V$.

I am not sure my answer is correct:

$\forall \alpha, \beta\in V$ the following holds (since we are under $\mathbb{R}$): $<T\alpha,\beta>\ =\ <(\alpha -\frac{2<v,\alpha>}{||v||^2}v),\beta> \ =\ <\alpha,\beta>-\frac{2<v,\alpha>}{||v||^2}<v,\beta>$

Let $\beta=v$. We get: $ <T\alpha, v>\ =\ <\alpha,v>-\frac{2<v,\alpha>}{||v||^2}||v||^2 \ = \ <\alpha,v>-2<\alpha,v> \ = \ -<\alpha,v> \ = \ <\alpha,-v> \implies <T\alpha, v>= <\alpha, -v> $ Now since $v$ is some fixed vector we get: $<\alpha, T^*v>=<\alpha, -v> \implies <\alpha, T^*v>-<\alpha, -v>=0 \implies <\alpha, T^*v+v>=0$ and since it is true for all $\alpha\in V$, it implies $ T^*v=-v$.

So, is it true that $T^*=-I?$ Thanks in advance :)

Answer 1

Let's compute

$$\begin{align} \langle Tx,y\rangle &= \left\langle x -\frac{2\langle v,x\rangle}{\|v\|^2} v, y\right\rangle \\ &= \langle x,y\rangle\color{blue}{-}\frac{\color{blue}{2}\langle \color{darkgreen}{v},\color{red}{x}\rangle\color{blue}{\langle v,y\rangle}}{\color{blue}{\|v\|^2}}\\ &= \langle x,y\rangle + \left\langle\color{blue}{ -\frac{2\langle v,y\rangle}{\|v\|^2}}\color{darkgreen}{v},\color{red}{x}\right\rangle \\ &= \left\langle x,y -\frac{2\langle v,y\rangle}{\|v\|^2}v \right\rangle,\end{align} $$and so $$T^*y = y -\frac{2\langle v,y\rangle}{\|v\|^2}v. $$Hence $T=T^*$.