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Let $\beta$ and ${\beta }^{\prime }$ be bases for the finite dimensional vector space $V$ of dimension n over the field $\mathbb{F}$, and let $Q=\left[{I}_{V}{\right]}_{{\beta }^{\prime }}^{\beta }$, where ${I}_{V}$ is the identity operator on $V$. I just recently proved that $Q\left[x{\right]}_{{\beta }^{\prime }}=\left[x{\right]}_{\beta }$ for every $x\in V$ (twas rather simple), which suggests the title of "basis transformation" matrix or "coordinate transformation" matrix for the matrix $Q$. I am now wondering whether the converse holds.
Let ${V}^{\prime }$ denote $V$ with its elements written with respect to the basis ${\beta }^{\prime }$, and suppose that $Q\in {M}_{n}\left(\mathbb{F}\right)$ is such that $Q\left[x{\right]}_{{\beta }^{\prime }}=\left[x{\right]}_{\beta }$ for every $x$. Since $\mathcal{L}\left({V}^{\prime }V\right)$ is isomorphic to the matrix algebra ${M}_{n}\left(\mathbb{F}\right)$ by sending a linear operator to its matrix representation, given $Q$ there exists a linear operator $T\in \mathcal{L}\left({V}^{\prime },V\right)$ such that $Q=\left[T{\right]}_{{\beta }^{\prime }}^{\beta }$. Hence $\left[T{\right]}_{{\beta }^{\prime }}^{\beta }\left[x{\right]}_{{\beta }^{\prime }}=\left[x{\right]}_{\beta }$ or $\left[T\left(x\right){\right]}_{\beta }=\left[x{\right]}_{\beta }$...
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.
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Kaya Kemp
Your equation $\left[T\left(x\right){\right]}_{\beta }=\left[x{\right]}_{\beta }$ says that $T\left(x\right)$ and $x$ have the same coordinates in the basis $\beta$. This means they can be written as the same linear combination of vectors of $\beta$ and thus $T\left(x\right)=x$ for all $x\in V$, i.e. $T={I}_{V}$.
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