structural:scholar_edition, information:filtering, linguistic:paraphrasing, information:extraction
First proof

Let V, W {\displaystyle V,W} be vector spaces over some field F, {\displaystyle F,} and T {\displaystyle T} defined as in the statement of the theorem with dim V = n {\displaystyle \dim V=n}.

As Ker T ⊂ V {\displaystyle \operatorname {Ker} T\subset V} is a subspace, there exists a basis for it. Suppose dim Ker T = k {\displaystyle \dim \operatorname {Ker} T=k} and let

be such a basis.

We may now, by the Steinitz exchange lemma, extend K {\displaystyle {\mathcal {K}}} with n − k {\displaystyle n-k} linearly independent vectors w 1, …, w n − k {\displaystyle w_{1},\ldots,w_{n-k}} to form a full basis of V {\displaystyle V}.

Let

such that

is a basis for V {\displaystyle V}.From this, we know that

We now claim that T (S) {\displaystyle T({\mathcal {S}})} is a basis for Im T {\displaystyle \operatorname {Im} T}.The above equality already states that T (S) {\displaystyle T({\mathcal {S}})} is a generating set for Im T {\displaystyle \operatorname {Im} T}; it remains to be shown that it is also linearly independent to conclude that it is a basis.

Suppose T (S) {\displaystyle T({\mathcal {S}})} is not linearly independent, and let

for some α j ∈ F {\displaystyle \alpha _{j}\in F}.

Thus, owing to the linearity of T {\displaystyle T}, it follows that

This is a contradiction to B {\displaystyle {\mathcal {B}}} being a basis, unless all α j {\displaystyle \alpha _{j}} are equal to zero. This shows that T (S) {\displaystyle T({\mathcal {S}})} is linearly independent, and more specifically that it is a basis for Im T {\displaystyle \operatorname {Im} T}.

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

Finally we may state that

This concludes our proof.