Fix $u,x \in V$ with $u\ne 0$. Define $T\in \mathcal L(V)$ by $$Tv = \langle v, u\rangle x$$ for every $v\in V$. Prove that $$\sqrt{T^*T}v = \frac{\|x\|}{\|u\|}\langle v,u\rangle u$$ for every $v\in V$.
I try to use $\|Tv\|^2 = \langle Tv, Tv\rangle = \langle T^*Tv, v\rangle$ to relate this to $\sqrt{T^*T}$, but I got stuck. Can anyone help me?
Assuming you're dealing with real vector spaces, you have, for any $v,w\in V$, $$ \langle T^*Tv,w\rangle= \langle Tv,Tw\rangle= \langle \langle v,u\rangle x,\langle w,u\rangle x\rangle= \langle v,u\rangle\langle w,u\rangle\langle x,x\rangle= \langle \langle v,u\rangle\langle x,x\rangle u,w\rangle $$ so $$ T^*Tv=\langle v,u\rangle\langle x,x\rangle u $$ Now, the map defined by $$ S(v) = \frac{\|x\|}{\|u\|}\langle v,u\rangle u $$ is self-adjoint (easy) and positive semidefinite because $$ \langle v,S(v)\rangle= \left\langle v,\frac{\|x\|}{\|u\|}\langle v,u\rangle u\right\rangle= \frac{\|x\|}{\|u\|}\langle v,u\rangle^2\ge0 $$ Moreover $$ S(S(v))=\frac{\|x\|}{\|u\|}\langle S(v),u\rangle u= \frac{\|x\|}{\|u\|}\left\langle \frac{\|x\|}{\|u\|}\langle v,u\rangle u,u\right\rangle u= \frac{\langle x,x\rangle}{\langle u,u\rangle} \langle v,u\rangle\langle u,u\rangle u $$ so $S^2=T^*T$.