Dimension and base of a subspace

Question

In the $Q$ field of rational numbers, the subspace of $Q^5$ is

$V = {(a, b, c, d, e) ∈ Q^5 : b = 0, a + c = d + e}$

How can I find the dimension of V and its base?

Answer 1

Well, since $b=0$, it does not contribute to dimension.

So, $V \cong V^{\prime}=\{(a,c,d,e) \mid a+c=d+e\}$.

The latter condition is the kernel of the map $\mathbb Q^4 \to \mathbb Q$ given by $(a,c,d,e) \mapsto a+c-d-e$.

Is the map surjective? What is the dimension of its kernel?

As a last ditch resort, write down the matrix and find its rank (or equivalently the dimension of its nullspace.) But this is more work than it is worth.