# Extending a functional on a subspace to one on the whole vector space

The dual space of a vector space $V$ is the vector space of all linear functionals on $V$. Denote the dual space of $V$ by $V'$.

Question: If $W$ is a subspace of a finite dimensional vector space $V$ and $f\in W'$, then there exists $g\in V'$ such that $f=g$ on $W$.

Any help on this please, I have no idea how to start. Thanks.

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You mean $V$ is a finite-dimensional space. – 1015 Feb 22 '13 at 12:59
This is true if $V$ has infinite dimension also. – 1015 Feb 22 '13 at 13:16

There exists a complement $U$ for $W$ in $V$: $$W\oplus U=V.$$ Recall this means that every $v\in V$ can be uniquely written $v=w+u$ with $w\in W$ and $u\in U$.

This can be seen by taking a basis of $W$ and then completing it into a basis of $V$.

Now extend $f$ to $V$ by $0$, for instance: $$g:v=w+u\longmapsto f(w).$$ Then $g=f$ on $W$.

Note: the same proof works in infinite dimension. Related is the Hahn-Banach theorem which allows the extension of bounded linear functionals to bounded linear functionals on a normed vector space: http://en.wikipedia.org/wiki/Hahn%E2%80%93Banach_theorem

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Ugly way: choose a basis $B_0$ for $W$, and choose a basis $B$ for $V$ such that $B_0 \subseteq B$. Set $g=f$ on $W$, and set $g(v)$ arbitrarily for $v \in B \setminus B_0$. Extend by linearity.

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Hint: If you start with any basis for $W$, you can extend it to a basis for $V$. And a linear functional is completely determined by a choice of its values on a basis.

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