Let ${V}^{\prime}$ denote $V$ with its elements written with respect to the basis ${\beta}^{\prime}$, and suppose that $Q\in {M}_{n}(\mathbb{F})$ is such that $Q[x{]}_{{\beta}^{\prime}}=[x{]}_{\beta}$ for every $x$. Since $\mathcal{L}({V}^{\prime}V)$ is isomorphic to the matrix algebra ${M}_{n}(\mathbb{F})$ by sending a linear operator to its matrix representation, given $Q$ there exists a linear operator $T\in \mathcal{L}({V}^{\prime},V)$ such that $Q=[T{]}_{{\beta}^{\prime}}^{\beta}$. Hence $[T{]}_{{\beta}^{\prime}}^{\beta}[x{]}_{{\beta}^{\prime}}=[x{]}_{\beta}$ or $[T(x){]}_{\beta}=[x{]}_{\beta}$...

I want to say that this implies $T={I}_{V}$, but I can't clearly see what lemma I need in order to make that conclusion.