Showing that $f_i(v) = \delta_{i,j}$ for some $v$ for a l.i. set on $V^*$

Question

Let $V$ be a linear space and $\alpha = f_1,\dots, f_m$ be a l.i. family in $V^*$(the dual space of $V$). Show that for every $1\le j\le m$ there exists a $v\in V$ such that $f_i(v) = \delta_{i,j}$ for all $1\le i\le m$

I understand that not all $f_i(v)$ for all $v$ can be zero if they're l.i. and that given that $f_i(w)=a\ne 0$ for some $w$ I can make it be $1$ for some $v$ by taking $v=\frac{1}{a}w$ but I'm stuck with how to show that all other $i$ are zero for some $v$. Any help would be appreciated.

Rodrigo Dias · Answer 1 · 2017-10-04T02:30:35.303

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Here is an intuitive idea: think hyperplanes!!!

Since $\{f_j\}_{j=1}^m$ is linearly independent, we know that $\ker f_i \neq \ker f_j$ for $i\neq j$. Besides, $\dim\ker f_i = \dim V -1$ for all $i\in\{1,\cdots,m\}$.

Let $k\in\{1,\cdots, m\}$ be fixed. We shall construct $v_k\in V$ such that $f_i(v_k) = \delta_{ik}$. One knows that $W = \bigcap_{i\neq k} \ker f_i$ has dimension $\dim V - m+1>0$. Now, we are allowed to take $w_k\in W$ such that $f_k(w_k) = a_k \neq 0$ (and thus define $v_k = \frac{1}{a_k} w_k$, as you thought) because $\{f_j\}$ is linearly independent.

edited Oct 04 '17 at 02:30

answered Oct 04 '17 at 02:15

Rodrigo Dias

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Sorry, I can't see why $\ker f_i \ne \ker f_j$. What is $n$? – AnalyticHarmony Oct 04 '17 at 02:25
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Sorry, $n = \dim V$, I've edited it. For your question, see this https://math.stackexchange.com/q/890101/375952 – Rodrigo Dias Oct 04 '17 at 02:37
But I can't assume dimension is finite – AnalyticHarmony Oct 04 '17 at 02:38
Right, my bad. But it's sufficient to know that such $W$ isn't the null space, and it follows from the fact that $f_j\neq 0$ for all $j$, can you see it? – Rodrigo Dias Oct 04 '17 at 02:43

Showing that $f_i(v) = \delta_{i,j}$ for some $v$ for a l.i. set on $V^*$

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