Bounded operators on complex Banach space $X$ are commutative exactly when $X$ is one-dimensional? I am trying to prove that for a Banach space $X$ over $\mathbb{C}$, dim$(X)=1$ if and only if $\mathfrak{B}(X)$ is commutative.
From this StackExchange question (Bounded linear operator commuting with every compact operators), we can see that $A \in \mathfrak{B}(X)$ commuting with each $K \in \mathfrak{K}(X)$ (the space of compact operators) means that $A = \lambda I$ for some scalar $\lambda \in \mathbb{C}$.
I can convince myself that the result follows (every bounded operator being commutative means that they commute with compact operators, so they are of the form $\lambda I$), but I don't know how to rigorously get to the conclusion that dim$(X)=1$, let alone how one would show the other direction.
Any hint or help is highly appreciated.
 A: The direction $\dim(X)=1\implies\mathcal{B}(X)\text{ is commutative}$ is trivial, since every linear map $\mathbb{C}\to\mathbb{C}$ is of the form $z\mapsto c\cdot z$, where $c\in\mathbb{C}$ is a constant.
For the converse: We show that if $\dim(X)>1$ then we may find two operators that do not commute. Take a basis of $X$, let's say $E$. Since $\dim(X)>1$, $E$ has at least two elements, say $x_1,x_2\in E$. By the Hahn-Banach theorem we may find a functional $\phi\in X^*$ such that $\phi(x_1)=1$, $\phi(x_2)=0$ and a functional $\psi\in X^*$ such that $\psi(x_1)=0$, $\psi(x_2)=1$.We now set $T:X\to X$ by $Tx=\phi(x)\cdot x_1$ and $S:X\to X$ by $Sx=\psi(x)x_1$. These are bounded because the functionals are bounded.
Note that $TSx=T(\psi(x)x_1)=\psi(x)Tx_1=\psi(x)\phi(x_1)x_1=\psi(x)x_1=Sx$, i.e. $TS=S$. On the other hand, $STx=S(\phi(x)x_1)=\phi(x)S(x_1)=\phi(x)\psi(x_1)x_1=0$, i.e. $ST=0$. Since $S\neq0$ we have that $TS\neq ST$.
Note: Hahn-Banach was necessary. If we simply defined the operators on the basis $E$ and then extended linearly, then it is not apparent why the operators are bounded. Unless the case is $\dim(X)=n<\infty$, where everything is bounded.
