# 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.

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.
• You do not need a basis $E$, only two linearly independent vectors $x_1,x_2$.