# If $\ker f\subset \ker g$ where $f,g$ are non-zero linear functionals then show that $f=cg$ for some $c\in F$.

Let $$V$$ be a vector space with $$\dim V=n$$ .

If $$\ker f\subset \ker g$$ where $$f,g$$ are non-zero linear functionals then show that $$f=cg$$ for some $$c\in F$$.

Now let $$\mathcal B=\{v_1,v_2,\ldots ,v_n\}$$ be a basis of $$V$$,

since $$f,g$$ are non-zero linear functionals then $$\exists v_i\in \mathcal B$$ such that $$g(v_i)\neq 0\implies f(v_i)\neq 0$$

Take $$i=1$$ without any loss of generality so take $$g(v_1)\neq 0,f(v_1)\neq 0$$.

Now take $$c=\dfrac{f(v_1)}{g(v_1)}$$

Then we need to show that $$(f-cg)(v_i)=0\forall i$$

Now $$(f-cg)(v_1)=0$$

How to show that $$(f-cg)(v_i)=0\forall i\ge 2$$

Note::Another Question Why do we need the dimension of the vector space to be finite?

• When you say linear functionals, you mean that $f, g : V \rightarrow k$ (where $k$ is the field $V$ is over) right? Dec 23, 2018 at 10:40
• @AniruddhAgarwal,yes u r right
– user596656
Dec 23, 2018 at 10:40
• Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/… Dec 23, 2018 at 15:12

There is no need for finite dimensionality and no need to use bases. Let $$f(x) \neq 0$$, $$y$$ be arbitrary and consider $$y-\frac {f(y)} {f(x)} x$$. By linearity we get $$f(y-\frac {f(y)} {f(x)} x)=0$$. By hypothesis this implies $$g(y-\frac {f(y)} {f(x)} x)=0$$. Hence $$g(y)=cf(y)$$ where $$c=\frac {g(x)} {f(x)}$$. Hypothesis implies that $$c \neq 0$$ so we can write $$f=\frac 1 c g$$.

• Did you switch up $f$ and $g$? Dec 23, 2018 at 15:09
• @ShubhamJohri Thanks for the comment. I have corrected the answer. Dec 23, 2018 at 23:15

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).

By definition $$f,g:V\rightarrow\mathbb{R}$$ are non-zero linear functionals (you can replace $$\mathbb{R}$$ with $$\mathbb{C}$$ or any field). By the rank-nullity theorem we have that $$\dim \ker f = \dim \ker g = n-1$$ Since $$\ker f \subseteq \ker g$$ we conclude that $$\ker f = \ker g$$. (In some sense this shows that $$f-cg(v_i)=0$$ in your solution because $$f(v_i)=g(v_i)=0$$.)

Now take a basis $$v_1,...,v_{n-1}$$ for the kernel and take $$v$$ which is linearily independent of those. Then $$f(v),g(v)\not = 0$$ are real numbers.

Take $$c=\frac{f(v)}{g(v)}$$. Since $$f,g$$ are non-zero only on $$\text{span} ({v})$$ the rest of the claim is immediate

• Nice approach !it does not answer for the infinite dimensional case
– user596656
Dec 23, 2018 at 12:03