Proving adjoint operator identity $(I-\lambda T)^*=I^*- \bar{\lambda}T^* $

Question

Let $T\in \cal{B}$$(H,H)$, where $H$ is a Hilbert space and $\lambda \in \mathbb{C}$. I need to show that $(I-\lambda T)^*=I^*- \bar{\lambda}T^* $ .

My most likely wrong attempt:

$\langle y,(I-\lambda T)x\rangle=\overline{\langle(I-\lambda T)x,y\rangle}=\overline{\langle y,(I-\lambda T)^*x\rangle} ={\langle y,I^*x\rangle}-\overline{\lambda}\langle y,T^*x \rangle$.

I'm not sure if I'm allowed to do last step, couldnt find a similar thread so I'm asking for clarification.

Answer 1

\begin{align} \langle (\lambda I-T)^*x,y\rangle&=\langle x,(\lambda I-T)y\rangle \\ &= \langle x,\lambda y-Ty\rangle \\ &= \overline{\lambda}\langle x,y\rangle -\langle x,Ty\rangle \\ &= \overline{\lambda}\langle x,y\rangle - \langle T^*x,y\rangle \\ &= \langle (\overline{\lambda}I-T^*)x,y\rangle \end{align} Because this holds for all $y$, then $(\lambda I-T)^*x=(\overline{\lambda}I-T^*)x$ holds for all $x$, which gives $(\lambda I-T)^*=\overline{\lambda}I-T^*$.