Question

If Z is an (n — l)-dimension£il subspace of an n-dimensioned 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.

Answer 1

Answer on Question #40830 – Math – Functional Analysis

Question. If $Z$ is an $(n-1)$-dimensional subspace of an $n$-dimensioned 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.

Solution. Let $e_{1},\ldots,e_{n-1}$ be a basis for $Z$. Since $\dim X=n$, there exists a vector $e_{n}\in X$ such that the vectors

$e_{1},\ \ldots,\ e_{n-1},\ e_{n}$

constitute a basis for $X$.

Then each $x\in X$ can be uniquely represented as a linear combination of $\{e_{i}\}_{i=1}^{n}$, that is

$x=a_{1}e_{1}+\cdots+a_{n}e_{n}$

for a unique $n$-tuple of numbers $(a_{1},\ldots,a_{n})$ and these numbers are called the *coordinates* of $x$ in this basis. Also notice that $x=(a_{1},\ldots,a_{n})\in Z$ if and only if $a_{n}=0$.

Furthermore, every linear functional $f:X\to\mathbb{R}$ is uniquely determined by its values on basis vectors $e_{1},\ldots,e_{n}$. Indeed, denote $a_{i}=f(e_{i})$, then for any $x=(x_{1},\ldots,x_{n})=x_{1}e_{1}+\cdots+x_{n}e_{n}$,

$f(x)=f(x_{1},\ldots,x_{n})=f(x_{1}e_{1}+\cdots+x_{n}e_{n})=x_{1}f(e_{1})+\cdots+x_{n}f(e_{n})=a_{1}x_{1}+\cdots+a_{n}x_{n}.$

We should construct a linear functional $f:X\to\mathbb{R}$ such that $f(x)=0$ if and only if $x\in Z$. Since $e_{1},\ldots,e_{n-1}\in Z$ and $e_{n}\not\in Z$, it follows that

$f(e_{1})=\cdots=f(e_{n-1})=0$

and

$f(e_{n})\neq 0.$

Therefore

$f(x)=f(x_{1}e_{1}+\cdots+x_{n}e_{n})=x_{n}f(e_{n}),$

so $f$ is uniquely determined by its *non-zero* value of $e_{n}$.

Thus for any $a\in\mathbb{R}\setminus\{0\}$ the functional $f_{a}(x_{1},\ldots,x_{n})=ax_{n}$ has the required properties: $f(x)=0$ if and only if $x\in Z$. In particular, such $f$ exists.

Moreover, if $f_{b}(x_{1},\ldots,x_{n})=bx_{n}$ is another such linear functional with $b\in\mathbb{R}\setminus\{0\}$, then

$f_{b}(x)=bx_{n}=\tfrac{b}{a}\cdot ax_{n}=\tfrac{b}{a}\cdot f_{a}(x),$

and so they differs by constant multiple $\tfrac{b}{a}$.

Thus a linear functional $f$ on $X$ for which $Z$ is a null space exist and is uniquely determined to within a scalar multiple.