# Null Space of A linear Functional

• Mar 28th 2011, 11:55 PM
raed
Null Space of A linear Functional
Dear Colleagues,

If $\displaystyle Z$ is an $\displaystyle (n-1)-$dimensional subspace of an $\displaystyle n-$dimensional vector space $\displaystyle X$, show that $\displaystyle Z$ is the null space of a suitable linear functional $\displaystyle f$ on $\displaystyle X$, which is uniquely determined to within a scalar multiple.

Regards,

Raed.
• Mar 29th 2011, 12:07 AM
Drexel28
Quote:

Originally Posted by raed
Dear Colleagues,

If $\displaystyle Z$ is an $\displaystyle (n-1)-$dimensional subspace of an $\displaystyle n-$dimensional vector space $\displaystyle X$, show that $\displaystyle Z$ is the null space of a suitable linear functional $\displaystyle f$ on $\displaystyle X$, which is uniquely determined to within a scalar multiple.

Regards,

Raed.

Let $\displaystyle Z$ have a basis $\displaystyle \{x_1,\cdots,x_{n-1}\}$ and extend it to a basis $\displaystyle \{x_1,\cdots,x_n\}$ for $\displaystyle X$. Then, merely define $\displaystyle \varphi:X\to F$ by $\displaystyle x_k\mapsto \delta_{k,n}$ and extend by linearity.

Show then that $\displaystyle \ker\varphi= Z$ and moreover that the only other way to construct a linear functional was to extend the basis for $\displaystyle Z$ to a basis for $\displaystyle X$ by picking some other $\displaystyle x'_n\in \text{span}\{x_n\}$ which then amounts to any other such linear functional looking like $\displaystyle x_k\to \alpha\delta_{k,n}$ where $\displaystyle x'_n=\alpha x_n$. etc.

Now prove all of that
• Mar 29th 2011, 03:27 AM
tonio
Quote:

Originally Posted by raed
Dear Colleagues,

If $\displaystyle Z$ is an $\displaystyle (n-1)-$dimensional subspace of an $\displaystyle n-$dimensional vector space $\displaystyle X$, show that $\displaystyle Z$ is the null space of a suitable linear functional $\displaystyle f$ on $\displaystyle X$, which is uniquely determined to within a scalar multiple.

Regards,

Raed.

Let $\displaystyle \{x_1,\ldots ,x_{n-1}\}$ be a basis for $\displaystyle Z$ , and complete this to a basis $\displaystyle \{x_1,...,x_{n-1},x_n\}$ of

the whole n-dimensional space.

Now define $\displaystyle f:X\rightarrow \mathbb{F}\,,\,\,\mathbb{F}=$ the definition field, by

$\displaystyle f(x_i)=\left\{\begin{array}{ll}0&\mbox{ , if }i=1,...,n-1\\1&\mbox{ , if }i=n\end{array}\right.$ and extend the definition by linearity.

Show now that $\displaystyle Z=\ker f$

Tonio
• Mar 29th 2011, 04:02 AM
raed
Thank you very much.
• Mar 29th 2011, 04:20 AM
raed
How the extension by linearity can be done.

Regards.
• Mar 29th 2011, 04:32 AM
tonio
Quote:

Originally Posted by raed
How the extension by linearity can be done.

Regards.

This is a standar procedure: Just write any element of the space as a linear combination

of the basis and define $\displaystyle \displaystyle{f(v)=f\left(\sum\limits^n_{i=1}a_ix_ i\right):=\sum\limits^n_{i=1}a_if(x_i)$ ...

Tonio
• Mar 29th 2011, 04:57 AM
raed
I undetrstand you. Thank you very much.

Best Regards.