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

1. 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.

2. 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.

3. Thank you very much for your reply.

4. Originally Posted by raed Thank you very much for your reply.
Just as a point, you should press the 'thank button' located on the bottom left of the posts for which you appreciate.

ndimensional, proper, space, subspace, vector 