Spectrum for a bounded linear operator and its adjoint on a Banach space are same. 
I have to show that spectrum for a bounded linear operator and its adjoint on a Banach space are the same. Spectrum is defined as $$ \sigma(T)=\{\lambda\in \mathbb{K}\ :\ T-\lambda I \ \text{is invertible}.\} $$

I have to show $\sigma(T)=\sigma(T^*)$. Let $\lambda \notin \sigma(T)$; then $ (T-\lambda I ) $ is invertible and  bounded. This implies $(T-\lambda I)^*$ is also invertible,  since $$ (T^*-\lambda I)^{-1}=[(T-\lambda I)^*]^{-1}\implies T^*-\lambda I \ \text{is invertible}.
$$
So $\lambda\notin \sigma(T^*).$
I am unable to prove the other part. Can anyone help me please?
Thanks.

$T-\lambda I$ is invertible if and only if $(T-\lambda I)^*=T^*-\lambda I$ is invertible: Since for every linear operator $A$ invertibility of $A$ and of $A^*$ are equivalent, which follows by taking the adjoints of, e.g., $AA^{-1}=I$ and $A^{-1}A=I$.
