kramberol

2022-07-08

Let $\beta $ and ${\beta}^{\prime}$ be bases for the finite dimensional vector space $V$ of dimension n over the field $\mathbb{F}$, and let $Q=[{I}_{V}{]}_{{\beta}^{\prime}}^{\beta}$, where ${I}_{V}$ is the identity operator on $V$. I just recently proved that $Q[x{]}_{{\beta}^{\prime}}=[x{]}_{\beta}$ 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 $Q\in {M}_{n}(\mathbb{F})$ 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.

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.

Kaya Kemp

Beginner2022-07-09Added 18 answers

Your equation $[T(x){]}_{\beta}=[x{]}_{\beta}$ says that $T(x)$ 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 $T(x)=x$ for all $x\in V$, i.e. $T={I}_{V}$.

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