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How do we obtain a linear extension of a linear functional $f$ where $f$ is defined on some subspace $Z$ of $X$? I always see something like the following phrase, 'let $E$ be the set of all linear extensions $g$ of $f$...but how do we construct such a $g$?

Here is what I tried - To make things simple, consider $X = \mathbb{R}^2 = \{(x_1,x_2)\ | x_1,x_2 \in \mathbb{R}\}$ and $Z = \{(x_1,0)\}$. Suppose $f$ is a linear functional on $Z$. Define $g$ as a extension of $f$ such that $g:X\to \mathbb{R}$ and $g|_Z = f$.

Now I want to show $g$ is in fact linear. Let $a = (a_1,0) + (0,a_2)$ and $b = (b_1,0) + (0,b_2)$ for $a,b\in \mathbb{R}^2$. Then we have \begin{align} a + b & = (a_1,0) + (0,a_2) + (b_1,0) + (0,b_2) \\ & = (a_1+b_1,0) + (0,a_2+b_2). \end{align}

Now I want to show linearity of the extension $g$ so I apply it to $a+b$: \begin{align} g(a+b) & = g((a_1+b_1,0) + (0,a_2+b_2)) \\ & = \dots \end{align} Here is where I am stuck. If I could say that $f = 0$ on $X\setminus Z$, then I could replace $g$ with $f$ and using the linearity of $f$ I could show $g$ is linear. But all I know is that $f$ is defined on $Z$ so it seems I can't do this.

So how do I should $g$ is linear?

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g is not zero on the complement! that is a big mistake a lot of people make! In fact, your definition actually involves PICKING a basis of your subspace and then extending to all of $V$, if $V$ is your vector space! After that you define $g(b_i):= f(b_i)$ if $b_i$ basivector of $Z$ and zero else. Hence $g$ is not $0$ on $V\setminus Z$ but only on the span of the additional basis vectors.

This is one of the many cases where people forget where and how important choices can be!

if you in fact want to set $g$ zero on the set theoretic complement for a non zero $f$, it will not be linear! let $v \in V\setminus Z $ and $w \in V\setminus Z$ s.t. $0 \neq v+w \in Z$ and $f(v+w)\neq 0$ (those exist by taking an element $v\in V \setminus Z$ and setting $w=v-z$ for $0\neq z \in Z\setminus \ker(f)$) then $$g(v-w)=g(z)=f(z) \neq 0=0+0 =g(v) + g(w) $$ (assuming that $f$ is not the zero map)

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  • $\begingroup$ Ok thanks, I see the issue now with defining $g$ in such a way. Is defining $g(b_j) = f(b_i)$ another way of extending $f$ to the entire space? That is, instead of setting $g(b_j) = 0$ for when $b_j$ is not a basis vector of $Z$, we instead set it to $f(b_i)$ where $b_i$ is an arbitrary basis vector of $Z$ (if we had more dimensions in our subspace..in the example here we only have one dimension so $b_i$ would just be, say, $b_1 = (1,0)$.) $\endgroup$
    – sonicboom
    Commented Nov 2, 2018 at 15:30
  • $\begingroup$ sure, however, you will produce another nontrivial kernel somewhere outside of $Z$, hence nothing would really change (i.e. it would be equivalent to the previous definition. just with a different basis, to see this, consider the dimension formula (Except of course your target space has enough additional dimensions to fill with elements not in the image of $f$)) yea, or (1,1),(1,2),... $\endgroup$
    – Felix
    Commented Nov 2, 2018 at 15:32

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