Showing that is a normal operator Let $H$ is a Hilbert space
$I$ is unit operator, $T \in B(H)$ and $\lambda \in \mathbb C$
$T$ is normal operator $\Rightarrow$ $T-\lambda I$ is a normal operator too.
I could only write :
I must show that $(T-\lambda I)(T-\lambda I)^{\ast}=(T-\lambda I)^{\ast}(T-\lambda I)$
$TT^{\ast}=T^{\ast}T$
$I^{\ast}=I$
$(T-\lambda I)^{\ast}=T^{\ast}- \bar{\lambda}I$ 
(where $\ast$ means adjoint and $\bar{\lambda}I$ means complex conjugate.
I cannot continue. I really stuck
Thanks for any help
 A: You can also use the characterization that $A \in B(H)$ is normal if and only if $\|Ax\| = \|A^*x\|, \forall x \in H$.
For $x \in H$ we have
\begin{align}
\|(T - \lambda I)x\|^2 &= \langle Tx - \lambda x, Tx - \lambda x\rangle \\
&= \langle Tx,Tx\rangle - 2\operatorname{Re} \langle Tx, \lambda x\rangle + \langle \lambda x, \lambda x\rangle\\
&= \|Tx\|^2 - 2\operatorname{Re} \overline{\lambda}\langle Tx, x\rangle + |\lambda|^2\|x\|^2\\
&= \|T^*x\|^2 - 2\operatorname{Re} \overline{\lambda}\langle x, T^*x\rangle + \left|\overline{\lambda}\right|^2\|x\|^2\\
&= \langle \overline{\lambda} x,\overline{\lambda}x\rangle - 2\operatorname{Re} \langle \overline{\lambda}x, T^*x\rangle+ \langle T^*x,T^*x\rangle\\
&= \left\langle \overline{\lambda}x - T^*x,\overline{\lambda}x - T^*x\right\rangle\\
&= \left\|\left(\overline{\lambda} I - T^*\right)x\right\|^2\\
&= \left\|(T-\lambda I)^*x\right\|^2
\end{align}
so $T - \lambda I$ is normal.
A: Then
$$(T-\lambda I)(T-\lambda I)^*=(T-\lambda I)(T^*-\overline\lambda I)
=TT^*-\lambda T^*-\overline\lambda T+|\lambda|^2I.$$
What about $(T-\lambda I)^*(T-\lambda I)$?
A: Note that
$(T - \lambda I)^\ast = T^\ast - \bar \lambda I; \tag 1$
then, using $TT^\ast = T^\ast T$ (i.e., the normality of $T$), one has
$(T - \lambda I)(T - \lambda I)^\ast = (T - \lambda I)(T^\ast - \bar \lambda I) = TT^\ast - \bar \lambda T - \lambda T^\ast + \lambda \bar \lambda I$
$= T^\ast T - \bar \lambda T - \lambda T^\ast + \lambda \bar \lambda I =  T^\ast (T - \lambda I) - \bar \lambda (T - \lambda I)$
$= (T^\ast - \bar \lambda I)(T - \lambda I) = (T - \lambda I)^\ast (T - \lambda I), \tag 2$
which asserts that $T - \lambda$ is normal.
A: $T$ is normal iff $T$ commutes with its adjoint $T^*$. If $T$ is normal, then $T^*$ commutes with $(T-\lambda I)$ and $\overline{\lambda}I$ commutes with $(T-\lambda I)$; hence $T^*-\overline{\lambda}I=(T-\lambda I)^*$ commutes with $T-\lambda I$, which makes $T-\lambda I$ normal.
