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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 $f(x) geq 0.$

One Answer

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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