Question regarding $H^0$ of stable vector bundle Let $M$ be a compact Riemann surface of genus $g$ and $E$ be a holomorphic vector bundle with the property that for any proper holomorphic subbundle $F$ we have $deg(F)/rank(F) < deg(E)/rank(E)$ then E is called a stable holomorphic vector bundle.

Hitchin states in one of his papers that $H^0(M, End (E)) ≈ \mathbb C$ since E is stable.

How does one prove this statement?
I'm new to this subject and I couldn't find a way to attempt this.
Any help is appreciated.
 A: Proposition
Let $E$ be a holomorphic vector bundle over $M$, then $$E \ \text{stable}\Longrightarrow E \ \text{simple}\Longrightarrow E \ \text{indecomposable}$$ where a vector bundle $E$ is simple if $End(E)\cong\mathbb C$ and it is indecomposable if can not be written as a direct sum of proper subbundles.
Proof
Let $E$ be a stable bundle and let $f\in \ $End$(E)$ be a non-zero homomorphism then it has to be necessarily an isomorphism (this result is quite standard in the theory of stable bundles). Thus End$(E)$ is a field, which contains $\mathbb{C}$ as its sub-field of scalar endomorphism. Then for any $f\in \ $End$(E)$, the sub-field $\mathbb{C}(f)\subset \ $End$(E)$ is a commutative field, and the Cayley-Hamilton Theorem implies that $f$ is algebraic over $\mathbb{C}$. Since $\mathbb{C}$ is algebraically closed, this shows that $f\in\mathbb{C}$. Hence End$(E)\simeq\mathbb{C}$, in paricular Aut$(E)\simeq\mathbb{C}^{*}$. Moreover if $E\simeq E_1\oplus E_2$ then Aut$(E)$ contains $\mathbb{C}^{*}\times\mathbb{C}^{*}$ and thus $E$ can not be simple, neither stable.
With this proposition in mind, it should be clear why $H^0(M,End(E))\cong\mathbb C$ if $E$ is a stable bundle.
