The dual basis spans the vector space of linear transformations.

Question

Proof attempt:

Let $T \in L(V,\Bbb{F})$ where $L(V,\Bbb{F})$ is the space of linear functionals of $V$ and where $\Bbb{F}$ is some field. Let $\mathscr{F}=\{\Phi_1,\cdots \Phi_n\}$ be the dual basis of, $B_v=\{v_1,\cdots, v_n\}$, the basis of $V$. Note $T(v)=T(\sum_{i=1}^{n}c_i v_i)=\sum_{i=1}^{n}c_i T(v_i)$ such that $\forall i, T(v_i)\in\Bbb{F}$. Since $T(v_i)\in\Bbb{F}$ implies $\exists \lambda_i\in\Bbb{F}, T(v_i)=\lambda_i=\lambda_i\cdot 1=\lambda_i\cdot\Phi(v_i)$. Then $T(v)=\sum_{i=1}^{n}(c_i\lambda_i)\Phi(v_i)$ and $T=\sum_{i=1}^n(c_i\lambda_i)\Phi_i$, as desired.

Answer 1

Since you are asking for criticism, a couple of (nit picky) remarks:

• The name 'dual basis' suggest that this set already spans the space of linear functionals. Perhaps it would be better to give a definition of $\{\Phi_i\}_i$ and just assert that it spans $L(V,\mathbb{F})$.
• In the last two sentences, it should be $\Phi_i(v_i)$.

Other than that, it seems fine.

Another approach, although ad-hoc, is to just claim that $T = \sum_{i = 1}^n T(v_i)\Phi_i$, since given $v = \sum_ic_iv_i$ we have that

$$ \begin{align} T(v) &= \sum_ic_iT(v_i) = \sum_ic_iT(v_i) \cdot 1 = \sum_ic_iT(v_i) \Phi_i(v_i)\\ &= \sum_iT(v_i) \cdot \Phi(c_iv_i) = \sum_{i} T(v_i)\Phi_i(v). \end{align} $$

with the latter equality given by the fact that

$$ \Phi_i(v) = \sum_jc_j\Phi_i(v_j) = c_i\Phi_i(v_i) = \Phi_i(c_iv_i). $$