Let's say I have a vector space
$E = ${$(x,y)\in \mathbb{R^2} | x+y = 0$}.
I can thus express $E$ as {$x(1,-1) | x \in \mathbb R$}.
But now, what do I do? The definition of the dimension of a space is "cardinality of a basis of V over its base field", but I'm not sure what the "base field" is. Is the dimension simply 1 because there's only one vector? How do I solve these problems, generally? Thanks.
On
The subspace generated by the vector $(1,-1)$ has dimension $1$ and its basis is $B=\{(1,-1)\}$
You may consider any other basis such as $B^*=\{(-3,3)\}$ as long as you have a non-zero vector which spans your subspace.
On
$E$ is defined to be the kernel of the map $T : \Bbb R^2 \to \Bbb R$ sending $(x,y)$ to $x+y$.
So it fits into the exact sequence: $$0 \longrightarrow E \longrightarrow \Bbb R^2 \overset T \longrightarrow \Bbb R$$
We see that $T$ is surjective, since it has a right inverse $s : \Bbb R \to \Bbb R^2$ sending $x$ to $(x,0)$. So, our exact sequence becomes:
$$0 \longrightarrow E \longrightarrow \Bbb R^2 \overset T \longrightarrow \Bbb R \longrightarrow 0$$
Since $\dim$ is additive, we have $\dim E + \dim \Bbb R = \dim \Bbb R^2$, i.e. $\dim E + 1 = 2$, i.e. $\dim E = 1$.
Yes, $\dim E = 1$ because the set $\{(1,-1)\}$ is a basis for $E$.
Indeed, it is linearly independent since clearly $\lambda(1,-1) = 0 \implies \lambda = 0$, and it spans your space $E$ because every $(x,y) \in E$ is of the form $\lambda(1,-1)$ for some scalar $\lambda$.
The base field is the same field as for $\mathbb{R}^2$, namely $\mathbb{R}$.