# Thread:

1. ## Null Space of A linear Functional

Dear Colleagues,

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.

2. Originally Posted by raed
Dear Colleagues,

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

3. Originally Posted by raed
Dear Colleagues,

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

4. Thank you very much.

5. How the extension by linearity can be done.

Regards.

6. 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

7. I undetrstand you. Thank you very much.

Best Regards.
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