Extending linear function from a subspace to the whole (finite-dimensional) space

Question

Let $V$ be a finite-dimensional vector space over the field $F$ and let $W$ be a subspace of $V$. If $f$ is a linear functional on $W$, prove that there is a linear functional g on $V$ such that $g(\alpha)$ = $f(\alpha)$ for each $\alpha$ in the subspace $W$.

Choose an element $a$ not in $W$, decompose it into $v\in W$ and $u\in W^\perp$, then assign another value as you wish to $u$. Then we have a linear functional defined on $span{W,u}$. Since $V$ is of finite dimension, the process will terminate and you will get the final result. — Golbez, Mar 23 '14 at 05:05
Thank you, but what is, in your notation, the meaning of $W^I$ — ViKaN, Mar 23 '14 at 05:11
Sorry for my ignorance. When no interior product is defined, I cannot say $W^\perp$. So you can neglect my step of decomposing, and the functional defined on $span{W,a}$ is enough. — Golbez, Mar 23 '14 at 05:14
Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. — Martin Sleziak, Mar 23 '14 at 07:08

score 7 · Accepted Answer · answered Mar 23 '14 at 05:14

Choose a basis $w_1, w_2, \ldots, w_k$ for $W$, where $k = \dim W$. Extend this to a basis of $V$ by adding linearly independent vectors $v_{k + 1}, v_{k + 2}, \ldots, v_n$, where $n = \dim V$ and $v_j \notin W$ for $k < j \le n$. Set $g(w_j) = f(w_j)$ for $1 \le j \le k$ and set $g(v_j) = 0$ for $k < j \le n$. This defines $g$ on the basis $\{w_1, w_2, \ldots, w_k, v_{k + 1}, v_{k + 2}, \ldots, v_n \}$. Extend $g$ to all of $V$ by linearity: if $x \in V$ is given (uniquely!) by $x = \sum \alpha_j w_j + \sum \beta_k v_k$ then we set $g(x) = \sum \alpha_j g(w_j) + \sum \beta_k g(v_k)$. It is easy to see that $g$ so defined is a linear functional on $V$. Furthermore since in fact $g(x) = \sum \alpha_j g(w_j) = \sum \alpha_j f(w_j$) for $x \in W$, we see that $g$ and $f$ agree on W. QED.

Nota Bene: It is worth noting that the $g(v_k)$ may in fact be taken arbitrarily in the above construction; choosing $g(v_k) = 0$ is merely a convenience. End of Note.

Hope this helps. Cheerio,

and as always,

Fiat Lux!!!

score 2 · Answer 2 · answered Mar 23 '14 at 05:14

Let $\;\{u_1,...,u_k\}\; $ be a basis of $\;W\;$, and complete it to a basis $\;\{u_1,...,u_k,u_{k+1},..,u_n\}\;$ of V. Now define

$$g(u_i):=\begin{cases}f(u_i)&,\;\;1\le i\le k\\{}\\0&,\;\;k+1\le i\le n\end{cases}$$

and extend the definition by linearity.

score 0 · Answer 3 · answered Mar 23 '14 at 05:15

The linear functional on the subspace is defined by its values on the basis of the subspace. This basis can be constructed in a way that makes it subset of the basis of the whole space. Then how can you extend your functional to the whole space? Just define some values for your functional on the rest of the basis.

It is apparent that this choice of values on the whole basis is not unique. Thus the extension is not unique.

Extending linear function from a subspace to the whole (finite-dimensional) space

3 Answers3