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This question is in connection with this one: Show that $\langle v,w\rangle _1=c\langle v,w\rangle _2$ for some scalar $c$..

Let $V$ be a vector space. Let $f,g:V\to F$ such that $\ker f =\ker g$. Show that $f=cg$ for some scalar $c$.

If $f\equiv 0\implies g\equiv 0$, take $c=1$ we are done.

If $f\neq 0$ then $f(v_0)\neq 0\implies g(v_0)\neq 0$.

How can I prove $f=cg$ from here?

Jeremy Upsal
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    The question and answer/s to which you have provided the link answer your question. If not, can you explain what is the difficulty that you have? – AnyAD May 22 '18 at 04:12
  • @AnyAD: In fact the questions are substantially different, though similar, since no inner products are introduced here. Thus, this question generalizes the other one. Indeed, there are plenty of vector spaces which are not self-dual, so representing functionals by inner products is not always feasible. Cheers! – Robert Lewis May 22 '18 at 04:51
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    @Robert Lewis Thank you for that, I assumed incirrectly that this was an inner product space – AnyAD May 22 '18 at 05:59

1 Answers1

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Well, let's see . . . a smallish lemma might kick this off:

Smallish Lemma: Let $V$ be a vector space over the field $F$, and let

$\lambda:V \to F \tag 1$

be linear over $F$; if $\lambda \ne 0$, then

a.) There exists $w \in V$ with

$\lambda(w) = 1; \tag 2$

b.) For any $v \in V$,

$v = aw + y, \; a = \lambda(v) \in F, \; y \in \ker \lambda; \tag 3$

c.) $a$ and $y$ are unique.

Proof of Smallish Lemma: For (a), since $\lambda \ne 0$, there is some $x \in V$ with

$\lambda(x) \ne 0; \tag 4$

set

$w = (\lambda(x))^{-1} x; \tag 5$

then

$\lambda(w) = \lambda( (\lambda(x))^{-1} x) = (\lambda(x))^{-1} \lambda(x) = 1; \tag 6$

For (b): if $v \in V$, set $a = \lambda(v)$ and $y = v - aw$, with $w \in V$ as in (a); then

$\lambda(y) = \lambda(v) - a \lambda(w) = \lambda(v) - \lambda(v) \cdot 1 = 0; \tag 7$

thus

$y \in \ker \lambda, \tag 8$

and

$v = aw + y = \lambda(v) w + y; \tag 9$

For (c), if $a_1, a_2 \in F$, $y_1, y_2 \in \ker \lambda \subset V$, and

$a_1 w + y_1 = v = a_2 w + y_2, \tag{10}$

then

$(a_1 - a_2)w = y_2 - y_1 \in \ker \lambda; \tag{11}$

so that

$a_1 - a_2 = (a_1 - a_2) \lambda(w) = \lambda((a_1 - a_2)w) = \lambda(y_2 - y_1) = 0; \tag{12}$

thus $a_1 = a_2$ from which (10) yields $y_1 = y_2$. End: Proof of Smallish Lemma.

We may apply the Smallish Lemma to the present situation as follows: if $f = 0$, take $c = 0$ and we are done. If $f \ne 0$, by Smallish there exists $w \in V$ with $f(w) = 1$; since for any $v \in V$,

$v = f(v)w + y, \; y \in \ker f = \ker g, \tag{13}$

$g(v) = g(f(v)w + y) = f(v)g(w) + g(y) = f(v)g(w) = g(w)f(v); \tag{14}$

now taking $c = g(w)$ we are done!

Robert Lewis
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