1. ## Linear Functionals

Dear Colleagues,

If $\displaystyle f_{1}, ....,f_{p}$ are linear functionals on an $\displaystyle n-$dimensional vector space $\displaystyle X$, where $\displaystyle p<n$, show that there is a vector space $\displaystyle x\neq 0$ in $\displaystyle X$ such that $\displaystyle f_{1}(x)=0,..., f_{p}(x)=0$. What consequences does this result have with respect to linear equations.

Regards,

Raed.

2. Originally Posted by raed
Dear Colleagues,

If $\displaystyle f_{1}, ....,f_{p}$ are linear functionals on an $\displaystyle n-$dimensional vector space $\displaystyle X$, where $\displaystyle p<n$, show that there is a vector space $\displaystyle x\neq 0$ in $\displaystyle X$ such that $\displaystyle f_{1}(x)=0,..., f_{p}(x)=0$. What consequences does this result have with respect to linear equations.

Regards,

Raed.
Merely note that since $\displaystyle p<n$ you have that $\displaystyle \text{span}\left\{f_1,\cdots,f_p\right\}\overset{\ text{de}\text{f.}}{=}V$ is such that $\displaystyle \dim_F V<\dim_F \text{Hom}\left(X,F\right)=n$. Thus, it follows that $\displaystyle \dim\text{Ann }V=n-\din_F V>0$ and so in particular there exists some non-zero $\displaystyle \Phi\in\text{Ann }V$ which by definition must satisfy $\displaystyle \Phi(f_k)=0$ for $\displaystyle k=1,\cdots,p$. But it is a common fact that for finite dimensional vector spaces every element $\displaystyle \Psi$ of $\displaystyle V^{\ast\ast}$ is of the form of some evaluation functional $\displaystyle f\mapsto f(x)$ for some fixed $\displaystyle x\in V$. Thus, in particular there exists some non-zero $\displaystyle v\in V$ such that $\displaystyle \Psi(f)=f(v)$ for every $\displaystyle f\in \text{Hom}(V,F)$ and so in particular $\displaystyle f_k(v)=\Phi(f)=0$ for $\displaystyle k=1,\cdots,p$.

3. Originally Posted by Drexel28
Merely note that since $\displaystyle p<n$ you have that $\displaystyle \text{span}\left\{f_1,\cdots,f_p\right\}\overset{\ text{de}\text{f.}}{=}V$ is such that $\displaystyle \dim_F V<\dim_F \text{Hom}\left(X,F\right)=n$. Thus, it follows that $\displaystyle \dim\text{Ann }V=n-\din_F V>0$ and so in particular there exists some non-zero $\displaystyle \Phi\in\text{Ann }V$ which by definition must satisfy $\displaystyle \Phi(f_k)=0$ for $\displaystyle k=1,\cdots,p$. But it is a common fact that for finite dimensional vector spaces every element $\displaystyle \Psi$ of $\displaystyle V^{\ast\ast}$ is of the form of some evaluation functional $\displaystyle f\mapsto f(x)$ for some fixed $\displaystyle x\in V$. Thus, in particular there exists some non-zero $\displaystyle v\in V$ such that $\displaystyle \Psi(f)=f(v)$ for every $\displaystyle f\in \text{Hom}(V,F)$ and so in particular $\displaystyle f_k(v)=\Phi(f)=0$ for $\displaystyle k=1,\cdots,p$.
Thank you very much for your reply. But what do you mean by $\displaystyle {Hom}(V,F)$ and $\displaystyle {Ann }V$.

4. Originally Posted by raed
Thank you very much for your reply. But what do you mean by $\displaystyle {Hom}(V,F)$ and $\displaystyle {Ann }V$.
Dual space and annihilator respectively.

5. In fact I am not able to understand the post. Could you please resolve the problem in another way if it is possible.

Regards.

6. Originally Posted by raed
In fact I am not able to understand the post. Could you please resolve the problem in another way if it is possible.

Regards.
No, I cannot. But you can.