# Let Z be a proper subspace of an n-dimensional vector space X

• Mar 29th 2011, 03:24 AM
raed
Let Z be a proper subspace of an n-dimensional vector space X
Dear Colleagues,

Let $\displaystyle Z$ be a proper subspace of an $\displaystyle n-$dimensional vector space $\displaystyle X$, and let $\displaystyle x_{0}\in X-Z$. Show that there is a linear functional $\displaystyle f$ on $\displaystyle X$ such that $\displaystyle f(x_{0})=1$ and $\displaystyle f(x)=0$ for all $\displaystyle x\in Z$.

Regards,

Raed.
• Mar 29th 2011, 06:25 AM
Drexel28
Quote:

Originally Posted by raed
Dear Colleagues,

Let $\displaystyle Z$ be a proper subspace of an $\displaystyle n-$dimensional vector space $\displaystyle X$, and let $\displaystyle x_{0}\in X-Z$. Show that there is a linear functional $\displaystyle f$ on $\displaystyle X$ such that $\displaystyle f(x_{0})=1$ and $\displaystyle f(x)=0$ for all $\displaystyle x\in Z$.

Regards,

Raed.

Of course this is true. Namely, for any vector spaces $\displaystyle V$ with basis $\displaystyle \{x_1,\cdots,x_n\}$ and vector space $\displaystyle W$ there exists a unique linear transformation $\displaystyle T$ such that $\displaystyle T(x_k)=w_k$ where $\displaystyle w_k\in W$ for $\displaystyle k=1,\cdots,n$. Indeed, just define $\displaystyle \displaystyle T\left(\sum_{r=1}^{n}\alpha_r x_r\right)=\sum_{r=1}^{n}\alpha_r w_r$. Clearly then this is a linear transformation which satisfies the condition and moreover it's clear that any linear transformation which satisfies that condition must look like that. Use this methodology here by noting that every linear functional $\displaystyle \varphi:V\to F$ is a linear transformation when $\displaystyle F$ is viewed as a one-dimensional vector space over itself.
• Mar 29th 2011, 07:35 AM
raed