I’m going through Linear Algebra Done Right, learning about singular value decomposition. I’m a bit lost when trying to calculate an adjoint of an operator. Obviously, this is simple when you are given an operator in matrix form. My question pertains to example 7.50:
Define operator $T$ on $F^4$ by $Tz = (0, 3z1, 2z2, 3z4)$. Find $T^*Tz$. ($T^*$ being the adjoint of operator $T$). How do you show $T^*Tz = (9z1, 4z3, 0, 9z4)$?
And exercise 7D.1:
Similarly, define operator $T$ on vector space $V$. Let $u$, $x$ belong to $V$, where $u \neq 0$. $T = \langle v, u\rangle x$. How do you show $T^*T = ||x||^2 \langle v, u\rangle u$?
Observe that for any $(w_1,w_2,w_3,w_4) \in \mathbf F^4$ we have $$ \begin{align} \langle T^*T(z_1,z_2,&z_3,z_4), (w_1,w_2,w_3,w_4) \rangle \\ &= \langle T(z_1,z_2,z_3,z_4), T(w_1,w_2,w_3,w_4) \rangle \\ &= \langle (0,3z_1,2z_2,-3z_4), (0,3w_1,2w_2,-3w_4) \rangle \\ &= 9z_1w_1 + 4z_2w_2 + 9z_4w_4 \\ &= \langle (9z_1,4z_2,0,9z_4), (w_1,w_2,w_3,w_4) \rangle \end{align} $$ and then $T^*T(z_1,z_2,z_3,z_4) = (9z_1,4z_2,0,9z_4)$.
On the other hand, for any $w \in V$ we have $$ \begin{align} \langle w,T^*Tv \rangle &= \langle Tw,Tv \rangle \\ &= \langle \langle w,u \rangle x, \langle v,u \rangle x \rangle \\ &= \langle w,u \rangle \langle v,u \rangle \|x\|^2 = \langle w,\overline{\langle v,u \rangle \|x\|^2}u \rangle \end{align} $$ and then $T^*Tv = \overline{\langle v,u \rangle \|x\|^2}u$.