Spectrum of non-invertible operator By an operator on Banach space $X$ we mean a bounded linear map $T:X\to X$. The spectrum of an operator $T$ on a complex Banach space $X$ is the set 
\begin{equation}
 \sigma(T)= \{\lambda\in\mathbb{C}: T-\lambda I \text{is not invertible}\}
\end{equation}
We denote by $\mathbb{D}$ the open unit disc  in the complex plane $\mathbb{C}$.
Can I say that:
1) If $\sigma(T)\subset \mathbb{D}$, then there is $0<t<1$ and $C\geq 0$  such that $||T^n||\leq Ct^n$ for all $n\in\mathbb{N}$?
2) If $T:X\to X$ is non-invertible, then
 $\sigma(T)\cap (\mathbb{C}-\overline{\mathbb{D}})=\emptyset$ ?
Please help me to know them.
 A: Yes to 1. If $\sigma(T)\subseteq\{ \lambda : |\lambda| < 1 \}$, then, because the spectrum is closed, there exists $0 < r < 1$ such that $\sigma(T)\subseteq \{ \lambda : |\lambda| \le r \}$, and there exists $N$ large enough that
$$
        \|T^n\|^{1/n} \le \frac{1+r}{2} < 1,\;\;\; n \ge N,\\
            \|T^n\| \le \left(\frac{1+r}{2}\right)^n,\;\; n \ge N.
$$
There is a constant $C > 1$ such that $\|T^n\| \le C\left(\frac{1+r}{2}\right)^n$ for all $1 \le n < N$. So $\|T^n\| \le C\left(\frac{1+r}{2}\right)^n$ for all $n \ge 1$.
No to 2. If $T$ is non-invertible, then the only thing you can say is that $0\in\sigma(T)$. The spectrum can be any compact subset of $\mathbb{C}$ that includes $0$, which would alow $\sigma(T)\cap(\mathbb{C}\setminus\mathbb{D})$ to be non-empty.
A: Hint: For 1) if the subset is not strict consider the identity operator on $\mathbb R^2$. If the subset is strict then remember we know that  $\rho(T)=\lim_{n\to\infty}\|T^n\|^{1/n}$, where $\rho(T)$ is the spectral radius, and we also know $\sigma(T)$ is compact.
For 2) consider the operator $T_A:\mathbb R^2\to \mathbb R^2$ given by the matrix: 
$$A=\begin{pmatrix}2&0\\0&0\end{pmatrix}.$$
