Proving a transformation from a vector space of functions is surjective

Question

Let $V$ be a finite dimensional vector space, and denote by $L(V)$ the set of all linear transformations from V to $\mathbb{R}$. Suppose that $\{\vec{v}_1 ... \vec{v}_n\}$ is a basis of $V$. Show that the map, $$T:L(V)\rightarrow \mathbb{R}_n, T: f\mapsto \begin{bmatrix}f(\vec{v}_1)\\.\\.\\.\\f(\vec{v}_n) \end{bmatrix} $$ is an isomorphism.

I think I'm alright on proving its injective, but I don't understand how to prove its surjective. I thought that I might use rank nullity theorem to prove that it had an $n$-dimensional image, but that doesn't work since the dimension of $L(V)$ is 1.

The other approach I had tried was to construct a basis for $\mathbb{R}^n$ through a set of functions, but I don't know how I would show it spans the target space. What's the correct approach to proving this?

Answer 1

Hint:

• A linear map $T:U\to V$ is completely determined by $T(\beta)$ where $\beta$ is a basis of $V$ and conversely, there exists a unique linear map $T\colon U\to V$for all $V_1\subseteq V$ such that $T(\beta)=V_1$ for some basis $\beta$ of $U$

• A linear map is bijective iff it carries basis, i.e., $T(\beta)$ is a basis for $V$ when $\beta$ is a basis of $U$.

Answer 2

Gven a vector $(a_1,\ldots,a_n)\in{\Bbb R}^n$, define a linear map from $V$ to $\Bbb R$ by $$f(x_1{\bf v}_1+\cdots+x_n{\bf v_n})=a_1x_1+\cdots+a_nx_n\ .$$ The map is well-defined (i.e. unambiguous) because if $${\bf v}=x_1{\bf v}_1+\cdots+x_n{\bf v_n}=y_1{\bf v}_1+\cdots+y_n{\bf v_n}$$ then $x_1=y_1$ etc. It is easy to show $f$ is linear. And it is clear from the definition that $T(f)=(a_1,\ldots,a_n)$.