# Linear Functional over $mathbb{R}$ Vector Space

Mathematics Asked by Saikat on October 21, 2020

If f and g are two non-zero linear functionals on $$mathbb{R}$$ vector space $$V$$[finite dimensional]. Such that whenever $$f(x)geq 0$$ we have $$g(x)geq 0$$. The claim is $$Ker(f)=Ker(g)$$ and $$f=alpha g$$ for some $$alpha > 0.$$

My Thinking:

Every linear functional over from $$V$$ to $$mathbb{R}$$ is of the form $$f(x)=sum_{i=1}^{n} a_ix_i$$ and $$g(x)=sum_{i=1}^{n} b_ix_i$$ where $$x=sum_{i=1}^{n} x_ie_i$$.
To conclude $$Ker(f)=Ker(g)$$.
$$f(x)=0$$ $$implies g(x) geq 0.$$ Now if $$g(x)=0$$ then nothing to show. If $$g(x) > 0$$ then considering
$$g(x)-f(x)=g(x)$$ we get $$sum_{i=1}^{n}(b_i-a_i)x_i=sum_{i=1}^{n}b_ix_i$$ We cannot conclude from here that $$b_i-a_i=a_i$$ for all $$1leq ileq n$$. Right? If no, then we can arrive at a contradiction that $$f$$ is $$0$$.

One Observation:

If $$(g(x) geq 0$$ and $$f(x)<0)$$ then $$f(-x)>0implies g(-x)geq 0 implies g(x)=0.$$

Hence if $$g(x) >0$$ then $$f(x) geq 0.$$

Let us assume to the contrary that there exists $$xinker(f)$$ such that $$xnotinker(g)$$. Now $$f(x)=0$$, so we must have $$g(x)ge0$$, or more precisely $$g(x)>0$$ as we have assumed $$g(x)ne0$$. But then $$g(-x)=-g(x)<0$$. Although $$f(-x)=0$$. This contradicts our condition. So $$ker(f)subseteqker(g)$$.

To prove the reverse inclusion, let us assume there exists $$xinker(g)$$ such that $$xnotinker(f)$$. Now take any $$y$$ such that $$g(y)<0$$. Such an $$y$$ must exist, because for any $$z$$ with $$g(z)>0$$, $$g(-z)<0$$, and we have assumed that $$g$$ is not the zero function. Now let $$k>|f(y)|/|f(x)|$$. Then if $$f(x)>0$$, then $$f(y+kx)>0$$, otherwise if $$f(x)<0$$, then $$f(y-kx)>0$$. But $$g(y+kx)=g(y-kx)=g(y)<0$$. Hence contradicts our condition. So $$ker(g)subseteqker(f)$$.

Take any $$xnotinker(g)$$. Let $$g(x)=r$$. Let $$y=r^{-1}x$$. Then $$g(y)=1$$. Also let $$alpha=f(y)$$. Now as $$f$$ and $$g$$ is linear non-zero functionals, so dimension of their image is $$1$$. By rank nullity theorem we have $$V=ker(g)+mathrm{span}(y)$$. So any $$vin V$$ can be written as $$v=z+ky$$, where $$zinker(g)$$ and $$kinmathbb{R}$$. Then $$f(v)=f(z+ky)=f(z)+kf(y)stackrel{(1)}{=}kalpha=alpha(g(z)+kg(y))=alpha g(v)$$ where $$(1)$$ follows from the fact that $$zinker(g)=ker(f)$$. Hence $$f=alpha g$$.

Correct answer by QED on October 21, 2020

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