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.
 A: We want to prove that if $X$ is a Banach space and $T^*\in B(X^*)$ is invertible, then $T$ is invertible in $B(X)$.
We go through a few steps.

*

*Note that $\operatorname{ran} T$ is closed. indeed, let $W$ be the inverse of $T^*$, and let $\{Tx_n\}$ be a Cauchy sequence. Then
\begin{align}
\|x_n-x_m\|
&=\sup\{|f(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\
&=\sup\{|T^*Wf\,(x_n-x_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\
&=\sup\{|(Wf)\,(Tx_n-Tx_m)|:\ f\in X^*,\ \|f\|=1\}\\ \ \\
&\leq\|Tx_n-Tx_m\|\,\sup\{\|Wf\|:\ f\in X^*,\ \|f\|=1\}\\ \ \\
&=\|W\|\,\|Tx_n-Tx_m\|.
\end{align}
So $\{x_n\}$ is Cauchy; there exists $x\in X$ with $x=\lim x_n$. As $T$ is bounded, $Tx=\lim Tx_n$, and $\operatorname{ran} T$ is closed.


*$T$ is injective. Indeed, if $Tx=0$, then for any $f\in X^*$ we have $f=T^*g$ (since $T^*$ is surjective). Then $$f(x)=T^*g(x)=g(Tx)=g(0)=0.$$ Thus $f(x)=0$ for all $f\in X^*$, and so $x=0$.


*$T$ is surjective. Indeed, if $y\in X\setminus \operatorname{ran} T$, using Hahn-Banach (and the fact that $\operatorname{ran} T$ is closed) there exists $g\in X^*$ with $g(y)=1$, $g(Tx)=0$ for all $x$. But then $0=g(Tx)=T^*g(x)$ for all $x$, and so $T^*g=0$. As $T^*$ is injective, $g=0$; this is a contradiction. So $X=\operatorname{ran} T$, and $T$ is surjetive.


*Finally, since $T$ is bijective and bounded, by the Inverse Mapping Theorem it is invertible. Or one can note that the estimate $\|x\|≤\|W\|\,\|Tx \|$ from the first part of the proof is the boundedness of $T^{-1}$.
A: $T-\lambda I$ 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 $A^{-1}A=I$.
