Let $B = \{δ_1,...,δ_n\}$ be subset of $D(V)$ and assume there exists vector $0_V \ne v ∈ V$ satisfying $δ_i(v) = 0$. Show $B$ is linearly dependent

Question

Let $n$ be a positive integer and let $V$ be a vector space of dimension $n$ over a field $F$. Let $B = \{δ_1,...,δ_n\}$ be a subset of $D(V)$ and assume that there exists a vector $0_V \ne v ∈ V$ satisfying $δ_i(v) = 0$ for all $0 ≤ i ≤ n$. Show that $B$ is linearly dependent.

I'm not sure how to go about this. Any solutions/hints are greatly appreciated.

Answer 1

In order to be linearly independent, $c_1=...=c_n=0$ when $\sum_{i=0}^n c_i \delta_i(v)=0$. Evaluating each $c_i \delta_i=c_i=0$ (based on the given condition that $\delta_i(v)=0$ $\forall v \in V$), we see that $B$ must be linearly independent.

Answer 2

Consider the map $f\colon V\to F^n$ defined by $$ f(x)=\begin{bmatrix} \delta_1(x)\\ \vdots\\ \delta_n(x) \end{bmatrix} $$ and identify $D(F^n)$ with the space of $1\times n$ matrices. Then the transpose of this map is $$ f^*\colon D(F^n)\to D(V) $$ defined by $$ f^*(\begin{bmatrix}\alpha_1&\dots&\alpha_n\end{bmatrix})= \alpha_1\delta_1+\dots+\alpha_n\delta_n $$ Since $f$ is not an isomorphism, because $f(v)=0$, neither $f^*$ is an isomorphism, so it is not injective.