What does "extend linearly" mean in linear algebra? The following proposition is from one of Gowers's articles:

Let $X$ be a vector space, and $x\in X$, $x\neq {\bf 0}$. Then there exists a linear map $g:X\to {\mathbb R}$ such that $g(x)\neq 0$.
The existence of this map can be proved as follows. Using the axiom of choice, one can show that the vector $x$ can be extended to a basis of $X$. Let $g(x)=1$, let $g(y)=0$ and extend linearly.

Here are my questions:

*

*What does "extend linearly" mean in the proof?

*I don't see any context about what $y$ is in his proof. So what is it here?

 A: First of all, I'm assuming $X$ is an $\mathbb{R}$-vector space, then $\mathbb{R}$ is an $\mathbb{R}$-vector space and everything makes sense (remember that linear maps only make sense between a) vector spaces b) defined over the same field).
$x\in X$ is a non-zero vector, hence $\{x\}$ is an linearly independent set in $X$. Therefore (axiom of choice), we can extend it to a basis $B$ of $X$. This means that there exists $B\subset X$ a basis for $X$ such that $x\in B$.
Now, to define a linear map $g:X\to \mathbb{R}$ we can define it on a basis, and then extend by linearity. In this case, we do it like this:
let $g(x)=1$ for the $x$ we picked above. Now let $g(y)=0$ for every $y\in B\setminus \{x\}$.
This is a function $g: B\to \mathbb{R}$. It can be extended to a linear map $\tilde{g}: X\to \mathbb{R}$ like this:
Write $B=\{e_i\}_{i\in I}$. Every vector $v\in X$ is a linear combination of the vectors in $B$: there exist unique scalars $a_i$ such that $v=\sum_{i=1}^n a_i e_i$.
Then define the map $\tilde{g}:X\to \mathbb{R}$ as $$\tilde{g}(v)=\sum_{i=1}^n a_i g(e_i)$$ 
It is a linear map (check!)
(In fact, it's what one usually does in linear algebra: if $B$ is a basis for a vector space $V$, then given any other vector space $W$, for any function $g:B\to W$ there exists a unique linear map (called linear extension) $\tilde{g}: V\to W$ such that $\tilde{g}|_B=g$. In this case, $W$ is $\mathbb{R}$ as an $\mathbb{R}$-vector space).
A: (1) If $x\neq 0$ is a vector in $X$, then there is a basis ${\cal B}$ of $X$ such that $x\in {\cal B}$ by the axiom of choice. Note that a linear functional $T:X\to \mathbb{R}$ is completely determined by its action on ${\cal B}$ (or, indeed, any basis of $X$). 
(2) If $y\in {\cal B}$ but $y\neq x$, then we define $T(y)=0$; we also define $T(x)\neq 0$. Note that $T$ extends linearly to a linear functional on $X$. More precisely, if ${\cal B}=\{y_i\}_{i\in I}$ where $I$ is an index set and if $x=\Sigma_{i\in I} c_iy_i$ where all but finitely many coefficients in this sum are $0$, then we can define (and, in fact, are forced to define) $T(x)=\Sigma_{i\in I} c_iT(y_i)$.
Exercise 1: Let $V$ be a finite dimensional vector space and let $W$ be an arbitrary vector space. If $(v_1,\dots,v_n)$ is a basis of $V$ and if $(w_1,\dots,w_n)$ is an arbitrary tuple of vectors in $W$, prove that there exists a unique linear map $T:V\to W$ such that $T(v_i)=w_i$ for all $1\leq i\leq n$. 
Exercise 2: Let $V$ be an arbitrary (real) vector space.
(a) Prove (using the axiom of choice) that $V$ has a basis ${\cal B}=\{y_i\}_{i\in I}$ (where $I$ is an index set).
(b) If $c_i\in \mathbb{R}$ for all $i\in I$, prove that there exists a unique linear functional $T:V\to \mathbb{R}$ such that $T(y_i)=c_i$ for all $i\in I$. 
Challenge: Let $X$ be a (real) vector space. We say that $X$ is a normed linear space if there is a function $X\to [0,\infty)$ with the following properties (for notational convenience, we denote the image of $x\in X$ under this function by $\left\|x\right\|$):


*

*$\left\|x\right\|=0$ if and only if $x=0$.

*$\left\|\alpha\cdot x\right\|=\left|\alpha\right|\left\|x\right\|$ for all $x\in X$ and all scalars $\alpha\in \mathbb{R}$.

*$\left\|x+y\right\|\leq \left\|x\right\|+\left\|y\right\|$ for all $x,y\in X$.


A linear functional $g:X\to\mathbb{R}$ is said to be a bounded linear functional if

$\left\|g\right\|=\text{sup}\{\frac{\left|g(x)\right|}{\left\|x\right\|}:x\in X, x\neq 0\}<\infty$.

If $g:X\to\mathbb{R}$ is a bounded linear functional, then the norm of $g$ is defined to be the number $\left\|g\right\|$ (above).
Prove the Hahn-Banach theorem:

Let $X$ be a normed linear space (over $\mathbb{R}$, if you prefer)
  and let $Y\subseteq X$ be a subspace. If $f:Y\to \mathbb{R}$ is a
  bounded linear functional, prove that there exists a bounded linear
  functional $F:X\to\mathbb{R}$ such that $F(y)=f(y)$ for all $y\in Y$
  and $\left\|F\right\|=\left\|f\right\|$.

I hope this helps!
