Null Space of A linear Functional

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March 28th 2011, 11:55 PM
raed

Null Space of A linear Functional

Dear Colleagues,

Could you please help me in solving the problem:
If $Z$ is an $(n-1)-$ dimensional subspace of an $n-$ dimensional vector space $X$ , show that $Z$ is the null space of a suitable linear functional $f$ on $X$ , which is uniquely determined to within a scalar multiple.

Regards,

Raed.
March 29th 2011, 12:07 AM
Drexel28

Quote:

Originally Posted by raed

Dear Colleagues,

Could you please help me in solving the problem:
If $Z$ is an $(n-1)-$ dimensional subspace of an $n-$ dimensional vector space $X$ , show that $Z$ is the null space of a suitable linear functional $f$ on $X$ , which is uniquely determined to within a scalar multiple.

Regards,

Raed.

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

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

Now prove all of that
March 29th 2011, 03:27 AM
tonio

Quote:

Originally Posted by raed

Dear Colleagues,

Could you please help me in solving the problem:
If $Z$ is an $(n-1)-$ dimensional subspace of an $n-$ dimensional vector space $X$ , show that $Z$ is the null space of a suitable linear functional $f$ on $X$ , which is uniquely determined to within a scalar multiple.

Regards,

Raed.

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

the whole n-dimensional space.

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

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

Show now that $Z=\ker f$

Tonio
March 29th 2011, 04:02 AM
raed

Thank you very much.
March 29th 2011, 04:20 AM
raed

How the extension by linearity can be done.

Regards.
March 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 $\displaystyle{f(v)=f\left(\sum\limits^n_{i=1}a_ix_ i\right):=\sum\limits^n_{i=1}a_if(x_i)$ ...

Tonio
March 29th 2011, 04:57 AM
raed

I undetrstand you. Thank you very much.

Best Regards.