Editing Rank–nullity theorem (section)

===First proof===
Let <math>V, W</math> be vector spaces over some field <math>F,</math> and <math>T</math> defined as in the statement of the theorem with <math>\dim V = n</math>.

As <math>\operatorname{Ker}T \subset V</math> is a [[Linear subspace|subspace]], there exists a basis for it. Suppose <math>\dim\operatorname{Ker}T = k</math> and let
<math display="block">\mathcal{K} := \{v_1, \ldots, v_k\} \subset \operatorname{Ker}(T)</math>
be such a basis.

We may now, by the [[Steinitz exchange lemma]], extend <math>\mathcal{K}</math> with <math>n-k</math> linearly independent vectors <math>w_1, \ldots, w_{n-k}</math> to form a full basis of <math>V</math>.

Let
<math display="block"> \mathcal{S} := \{w_1, \ldots, w_{n-k}\} \subset V \setminus \operatorname{Ker}(T) </math>
such that
<math display="block"> \mathcal{B} := \mathcal{K} \cup \mathcal{S} = \{v_1, \ldots, v_k, w_1, \ldots, w_{n-k}\} \subset V </math>
is a basis for <math>V</math>.
From this, we know that
<math display="block">\operatorname{Im} T = \operatorname{Span}T(\mathcal{B}) = \operatorname{Span}\{T(v_1), \ldots, T(v_k), T(w_1), \ldots, T(w_{n-k})\}</math>
::<math> = \operatorname{Span}\{T(w_1), \ldots, T(w_{n-k})\} = \operatorname{Span}T(\mathcal{S}) .</math>
We now claim that <math>T(\mathcal{S})</math> is a basis for <math>\operatorname{Im} T</math>.
The above equality already states that <math>T(\mathcal{S})</math> is a generating set for <math>\operatorname{Im} T</math>; it remains to be shown that it is also linearly independent to conclude that it is a basis.

Suppose <math>T(\mathcal{S})</math> is not linearly independent, and let
<math display="block"> \sum_{j=1}^{n-k} \alpha _j T(w_j) = 0_W </math>
for some <math>\alpha _j \in F</math>.

Thus, owing to the linearity of <math>T</math>, it follows that
<math display="block"> T \left(\sum_{j=1}^{n-k} \alpha _j w_j \right) = 0_W \implies \left(\sum_{j=1}^{n-k} \alpha _j w_j \right) \in \operatorname{Ker} T = \operatorname{Span} \mathcal{K} \subset V .</math>
This is a contradiction to <math>\mathcal{B}</math> being a basis, unless all <math>\alpha _j</math> are equal to zero. This shows that <math>T(\mathcal{S})</math> is linearly independent, and more specifically that it is a basis for <math>\operatorname{Im}T</math>.

To summarize, we have <math>\mathcal{K}</math>, a basis for <math>\operatorname{Ker}T</math>, and <math>T(\mathcal{S})</math>, a basis for <math>\operatorname{Im}T</math>.

Finally we may state that
<math display="block"> \operatorname{Rank}(T) + \operatorname{Nullity}(T) = \dim \operatorname{Im} T + \dim \operatorname{Ker}T</math>
::<math> = |T(\mathcal{S})| + |\mathcal{K}| = (n-k) + k = n = \dim V .</math>
This concludes our proof.