Question

(Linear extension) Let Z be a proper subspace of an n-dimensional
vector space X, and let f be a linear fimctional on Z Show that f can
be extended linearly to X, that is, there is a linear functioned f on X
such that f ̃|z =f

Answer 1

Answer on Question #40829 – Math – Functional Analysis

Question. Let $Z$ be a proper subspace of an $n$-dimensional vector space $X$, and let $f$ be a linear functional on $Z$. Show that $f$ can be extended linearly to $X$, that is, there is a linear functional $F$ on $X$ such that $F|_{Z}=f$.

Proof. Suppose $\dim Z=k<n$, and let $e_{1},\ldots,e_{k}$ be a basis for $Z$, that is a maximal collection of linearly independent vectors in $Z$. It is known that every basis of a subspace of a finite-dimensional vector space $X$ can be extended to a basis of all of $X$. So let us extend the above basis of $Z$ to a basis

$e_{1},\ldots,e_{k},\ e_{k+1},\ldots,e_{n}$

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

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

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

Using coordinates in the above basis define a function $F:X\to\mathbb{R}$ by the following formula:

$F(a_{1},\ldots,a_{k},a_{k+1},\ldots,a_{n})=f(a_{1},\ldots,a_{k}),$

in other owrds

$F(a_{1}e_{1}+\cdots+a_{k}e_{k}+a_{k+1}e_{k+1}+\cdots+a_{n}e_{n})=F(a_{1}e_{1}+\cdots+a_{k}e_{k}).$

We claim that $F$ is a linear functional such that $F|_{Z}=f$.

Indeed, let $x=(a_{1},\ldots,a_{n})$, $y=(b_{1},\ldots,b_{n})\in X$, and $s,t\in\mathbb{R}$. Then

$F(sx+ty)$ $=F(sa_{1}+tb_{1},\ldots,sa_{n}+tb_{n})$

$=f(sa_{1}+tb_{1},\ldots,sa_{k}+tb_{k})$

$=sf(a_{1},\ldots,a_{k})+tf(b_{1},\ldots,b_{k})$

$=sF(a_{1},\ldots,a_{n})+tF(b_{1},\ldots,b_{n})$

$=sF(x)+tF(y),$

so $F$ is a linear functional.

Moreover, if

$x=a_{1}e_{1}+\cdots+a_{k}e_{k}=(a_{1},\ldots,a_{k},\underbrace{0,\ldots,0}_{n-k})\in Z$

then

$F(x)=F(a_{1}e_{1}+\cdots+a_{k}e_{k})=f(a_{1}e_{1}+\cdots+a_{k}e_{k})=f(x),$

so $F|_{Z}=f(x)$.