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I stuck at the following linear algebra problem. Could you give me some hints?

Let $V$ be a vector space. Given $g,\,f_1,\,f_2,\,...,f_r\in V^*$, prove that $g\in \mathrm{span}\,(f_1,\,f_2,\,...,f_r)$ if and only if $\cap^{r}_{i=1}\mathrm{ker}\,(f_i)\subset\mathrm{ker}\,(g)$.

The "only if" part seems obvious: if $g\in \mathrm{span}\,(f_1,\,f_2,\,...,f_r)$, then there exits scalars $\alpha_1,...,\alpha_r$ such that $g=\Sigma^{r}_{i=1} \alpha_if_i$. Thus for $\forall v\in \cap^{r}_{i=1}\mathrm{ker}\,(f_i)$, we have $g(v)=\Sigma^{r}_{i=1} \alpha_if_i(v)=0$, which implies $v\in \mathrm{ker}\,(g)$.

But for the "if" part, I have trouble showing that if $\cap^{r}_{i=1}\mathrm{ker}\,(f_i)\subset\mathrm{ker}\,(g)$, then $g$ is in the span of $\{f_i\}_{i=1,...,r}$.

(This is a homework problem so hints or the key ideas are preferred. Thank you for your time.)

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    $\begingroup$ Are you already acquainted with quotient vector spaces? Does $$V/\left(\bigcap_{i=1}^r \ker f_i\right)$$ look like something you might have already seen, or frighteningly unknown? $\endgroup$
    – Daniel Fischer
    Sep 30, 2014 at 17:48
  • $\begingroup$ Yes, I have studied the definition of quotient vector space (and three isomorphism theorems about quotient vector space). But I didn't see how it connects with this problem. $\endgroup$
    – Sen
    Sep 30, 2014 at 17:57
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    $\begingroup$ Let $K$ be the scalar field. Consider the map $F\colon V\to K^r$ given by $F(v) = (f_1(v),\dotsc,f_r(v))$. Let $W = \operatorname{im} F$. Consider the induced map $\tilde{g}\colon W\to K$. $\endgroup$
    – Daniel Fischer
    Sep 30, 2014 at 18:00
  • $\begingroup$ @DanielFischer Did you mean that $\tilde g:(f_1(v),...,f_r(v))\mapsto\Sigma^r_{i=1}\alpha_if_i(v)$ for some fixed r-tuple $\{\alpha_1,...,\alpha_r\}\in \mathrm K^r$? So every $\tilde g \circ \mathrm F$ is in the span of $\{f_1,...,f_r\}$. And what I need to prove is that if for $\forall g \in \mathrm V^*$ such that $\cap^r_{i=1}\mathrm ker(f_i) \subset \mathrm ker(g)$, we have such g can be decomposed as $\tilde g \circ \mathrm F$. Did I understand it correct? Still I didn't see where the quotient vector space $\mathrm V/\cap^r_{i=1}\mathrm {ker}\,(f_i)$ is needed in the proof. $\endgroup$
    – Sen
    Oct 1, 2014 at 15:35
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    $\begingroup$ Yes, that is what I mean. You need the quotient space to see that the induced map $\tilde{g}$ exists. $W \cong V/\ker F = V/\bigcap_{i=1}^r \ker f_i$, and in the theory of quotient spaces, one proves that a linear $h\colon V \to U$ induces a linear $\tilde{h}\colon (V/N)\to U$ if and only if $N\subset \ker h$. One can of course prove that a $\tilde{g}$ with $g = \tilde{g} \circ F$ exists if and only if $\ker F \subset \ker g$ without referring to quotient spaces, but if you've never heard of quotient spaces, proving that would be non-obvious. $\endgroup$
    – Daniel Fischer
    Oct 1, 2014 at 15:46

1 Answer 1

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Hint: Consider the finite dimensional case. Let $e_1,\dots,e_n$ be a basis of $V$. It suffices to find $\alpha_i$ such that for all $k \in \{1,\dots,n\}$: $$ \sum_{i=1}^r \alpha_i f_i(e_k) = g(e_k) $$ That is, to solve the system $$ \pmatrix{ f_1(e_1) & f_2(e_1)&\cdots\\ f_1(e_2) & f_2(e_2)& \cdots \\ \vdots & &\ddots } \pmatrix{\alpha_1 \\ \vdots \\ \alpha_r} = \pmatrix{g(e_1)\\ \vdots \\ g(e_n)} $$ Under what conditions on $f_i$ and $g$ does the system have a solution? Try to extend this intuition.

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