Let $\mathcal A$ be a Banach algebra over $\mathbb{C}$, $\mathcal X$ a irreducible left $\mathcal A$-module. If $x,y \in \mathcal X$ are linearly independent, there exists an element $a\in\mathcal A$ such that $ax=x$ and $ay=0$.
Is it true? If it is true, how to prove?
Thanks a lot.
This is false. Consider $\mathcal{A} = \mathbb{C}$ as an algebra over $\mathbb{R}$, and let $\mathcal{X} = \mathcal{A}$, viewed as a (left) $\mathcal{A}$-module via left multiplication.
If $M$ is a non-zero submodule of $\mathcal{X}$ and $0 \neq m \in M$ then $a = (a/m)\cdot m \in M$ for all $a \in \mathcal{A}$ so $M = \mathcal{X}$. Hence $\mathcal{X}$ is simple.
Now $1$ and $i$ are elements in $\mathcal{X}$ which are linearly independent over the ground field $\mathbb{R}$, however if $ax = x$ then $a = 1$ and $ay = i \neq 0$.
The same example works whenever your simple module has endomorphism ring $D := End_{\mathcal{A}}(\mathcal{X})$ strictly bigger than the base field of the algebra.
However if you ask the same question with the stronger condition of linear independence over $D$, then the result is true. This follows from the Jacobson Density Theorem.