# Dual basis and kernel intersection

1. Jun 1, 2014

### mahler1

The problem statement, all variable
Let $\phi_1,...,\phi_n \in V^*$ all different from the zero functional. Prove that

$\{\phi_1,...,\phi_n\}$ is basis of $V^*$ if and only if $\bigcap_{i=1}^n Nu(\phi_i)={0}$.

The attempt at a solution.

For $→$: Let $\{v_1,...,v_n\}$ be a basis of $V$ such that $\phi_i(v_j)=δ_{ij}$. Now let $x \in \bigcap_{i=1}^n Nu(\phi_i)={0}$, as $x \in V$ then $x=α_1v_1+...+α_nv_n$.
By hypothesis $\phi_i(x)=0$ for all $1\leq i \leq n$. But this means $0=\phi_i(x)=\phi_i(α_1v_1+...+α_nv_n)=α_i$ for all $1\leq i \leq n$, it follows $x=0$.

I don't know how to show the other implication. For ←, in the exercise is given the suggestion: if $\{ψ_1,...,ψ_n\}$ is a basis of $V^*$, one can try to show that the matrix obtained from writing $\phi_1,...,\phi_n$ in coordinates with respect to the basis $\{ψ_1,...,ψ_n\}$, is an invertible matrix, I am stuck and I haven't got a clue on how to use this idea, I would like any hints or another idea to prove the remaining implication.

2. Jun 2, 2014

### pasmith

An alternative approach is to assume that $\Phi = \{\phi_1, \dots, \phi_n\}$ is not a basis for $V^{*}$, and exhibit a non-zero $v \in \bigcap_i \ker(\phi_i)$.

To that end: let $\Phi' \subset \Phi$ be a linearly-independent subset of $\Phi$ of maximum dimension. This $\Phi'$ can then be extended to a basis $B$ of $V^{*}$. Now consider the dual basis of $B$.