A question about functionals and dual space

Question

I have a question that I'm having trouble proving

If $f_1, f_2 ,..., f_n$ are linearly independent functionals in an $n$-dimensional vector space $V$ to it's scalar field $F$ does there always exists a base $x_1, x_2,..., x_n$ of V such that $$f_i(x_j)=\delta_{ij}=\begin{cases}1 \qquad i=j \\ 0 \qquad i \ne j \end{cases}$$

I know I should put my work here but I don't know how to prove it. It's an exam problem that I have in two days and I would really appreciate some help

Answer 1

Some steps for reach the result:

• Prove that $f_1,\dots,f_n$ is a basis for $V^*$, the space of all linear functions from $V$ to $\mathbf F$.
• For each $v \in V$ define $\operatorname{ev}(v) : V^* \to \mathbf F$ by $\operatorname{ev}(v)(\phi) = \phi(v)$, and prove that $\operatorname{ev}(v) \in V^{**}$, where $V^{**}$ is the space of all linear functions from $V^*$ to $\mathbf F$.
• Prove that if $v \in V \setminus \{0_V\}$ then there exists $\phi \in V^*$ such that $\phi(v) \neq 0$. Conclude that $\operatorname{ev} : V \to V^{**}$ is injective, and then, conclude that any $\varphi \in V^{**}$ is $\operatorname{ev}(v_\varphi)$ for some $v_\varphi \in V$.
• If $\varphi_1,\dots,\varphi_n \in V^{**}$ is the dual basis for $f_1,\dots,f_n$, then for each $i$ between $1$ and $n$ let $x_i \in V$ such that $\varphi_i = \operatorname{ev}(x_i)$, and prove that $x_1,\dots,x_n$ is the desired basis for $V$.

Answer 2

The kernel of each $f_i$ has dimension $n-1$. What is the minimal dimension of $$\bigcap_{i \neq j} \operatorname{ker}(f_i) \text{ ? }$$

Answer 3

$\textbf{Hint:}$ Since $f_{1}, \ldots, f_{n}$ are linearly independent and $V^{*}$ has dimension $n$, they form a basis for $V^{*}$. Now let $\Lambda_{1}, \ldots, \Lambda_{n}$ in $V^{**}$ be the dual basis of $f_{1}, \ldots, f_{n}$. For any $v \in V$, we can consider the "evaluation at $v$" linear functional: \begin{equation*} \begin{split} \text{ev}_{v}: & \hspace{0.1cm} V^{*} \rightarrow K \\ &\varphi \mapsto \varphi(v). \end{split} \end{equation*} The linear map that associates each $v \in V$ to $\text{ev}_{v}$ is an isomorphism between $V$ and $V^{**}$. In particular, $\Lambda_{1}, \ldots, \Lambda_{n}$ belong to the image of this linear map, so...

Answer 4

Let $y_1,y_2,\cdots,y_n$ be a basis of $X$. Then $A=[f_i(y_j)]$ must be invertible. If this were not the case, then there would exist scalars $\alpha_1,\alpha_2,\cdots,\alpha_n$ that are not all zero such that $$ [f_i(y_j)]\left[\begin{array}{c}\alpha_1\\ \alpha_2 \\ \vdots \\ \alpha_n \end{array}\right] = 0. $$ But that would imply that $\alpha_1 f_1 + \cdots + \alpha_n f_n$ vanishes on the basis $\{ y_1,y_2,\cdots,y_n\}$ and, hence, must be the $0$ functional, which is a contradiction. So, because $A$ is invertible, there is a linear combination $F$ of the $f_i$ such that $F(x_j)=\delta_{j,k}$. And this is true for every given $k=1,2,\cdots,n$.