I was reading Sheldon Axler. In the proof of
Suppose $T\in\mathcal{L}(V)$ is normal and $v\in V$ is an eigenvector of $T$ with eigenvalue $\lambda$. Then $v$ is also and eigenvector of $T^*$ with eigenvalue $\bar{\lambda}$ where $T^*$ is the adjoint of $T$ and $T$ is normal operator that is $TT^*=T^*T$.
there is this line that $$0=||(T-\lambda I)v||=||(T-\lambda I)^*v||=||(T^*-\bar{\lambda}I)v||$$
How we are getting this $||(T-\lambda I)^*v||=||(T^*-\bar{\lambda}I)v||$ ?
This is simply using the following properties of the adjoint: $$(A+B)^* = A^* + B^* \tag{1}$$ $$(\lambda A)^* = \overline{\lambda} A^* \tag{2}$$ Can you prove them?
Edit. Proof of $(2)$: $$\langle x, (\lambda A)^* y\rangle = \langle \lambda Ax,y\rangle = \lambda \langle Ax,y\rangle = \lambda \langle x, A^*y\rangle = \langle x, \overline{\lambda} A^* y\rangle$$ for all $x,y\in \mathcal H$, where $\mathcal H$ is the Hilbert space you're working with.