Dual spaces and dual basis

Question

If V is a finite dimensional vector space over the field F with dual space V* = Hom(V,F) . How to prove every ordered basis for V* is the dual basis for some basis for V ?

Answer 1

Hint: given a basis $(\xi_1,\dots,\xi_n)$ of $V^*$, it has a dual basis in $V^{**}$; use the canonical isomorphism $V\to V^{**}$.

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For any vector space $V$, define the map $\omega_V\colon V\to V^{**}$ by $\omega_V(x)=\hat{x}$ such that, for $\xi\in V^*$, $$ \hat{x}(\xi)=\xi(x) $$ is a well-defined linear map (verify). If $V$ is finite dimensional and $x\ne0$, there exists $\xi\in V^*$ such that $\hat{x}(\xi)=\xi(x)\ne0$ (just complete $x$ to a basis). Therefore $\omega_V$ is injective.

The existence of a dual basis entails that $\dim V=\dim V^*$ and therefore also $\dim V^*=\dim V^{**}$. Hence $\omega_V$ is an isomorphism.

In particular, if $\{\xi_1,\dots,\xi_n\}$ is a basis of $V^*$, there exists its dual basis in $V^{**}$ and we can assume it is of the form $\{\hat{x}_1,\dots,\hat{x}_n\}$ for some $x_1,\dots,x_n\in V$, because $\omega_V$ is an isomorphism.

Since we have chosen a dual basis, we have, by definition, $$ \hat{x}_i(\xi_j)=\delta_{ij} $$ (Kronecker delta), which amounts to saying that $$ \xi_j(x_i)=\delta_{ji} $$ and so $\{\xi_1,\dots,\xi_n\}$ is the dual basis of $\{x_1,\dots,x_n\}$.

Answer 2

Let $n$ be the dimension of $V$, and let $\{b^1,\ldots,b^n\}$ be an ordered basis for $V^*$. Let $\{e_1,\ldots,e_n\}$ be some basis for $V$, and let $\{e^1,\ldots,e^n\}$ be the dual basis for $V^*$, so $e^i(e_j)=\delta^i_j$. Then there exist coefficients $\{a_{ij}\}_{1\le i,j\le n}$ in the field over which the vector space is taken, such that $b^i=a_{ij}e^j$ for all $1\le i\le n$ (summation convention in force). Then $b^i(e_k)=a_{ij}e^j(e_k)=a_{ij}\delta^j_k=a_{ik}$. Since, for all $1\le i\le n$, there exists at least one $1\le k\le n$ such that $a_{ik}\neq0$, and this gives you the required basis for which $\{b^1,\ldots,b^n\}$ is the dual basis.