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**Deveno** um, the splitting lemma for vector spaces IS, in some sense, the rank-nullity theorem. which, in turn, IS the first isomorphism theorem for vector spaces.

L is just the null space of A, so dim(L) = nullity(A).

it's clear L = ker(A) (as considered as an element of Hom_F(F^m,F^n)). of course, one can just show L is a subspace directly:

suppose x,y are in L:

then A(x+y) = A(x) + A(y) (matrices are linear)

= 0 + 0 = 0 (these are 0-vectors in F^m), so x+y is in L.

similarly, if a is in F: A(ax) = a(A(x)) = a0 = 0, so ax is in L, whenever x is.

finally, L is always non-empty, since A(0) = 0, so at the very least, the 0-vector of F^n is in L.

by column-reducing the matrix A, we get a matrix AP, where P is an invertible matrix representing the product of elementary column-operation matrices.

if {e1,e2,....,en} is the standard basis for F^n, we have AP(ej) ≠ 0, for j = 1,2,..,r,

and AP(ej) = 0 for j = r+1,...,n. thus the vectors P(ej) for j = r+1,...,n are all in L, and linearly independent by the invertibility of P.

since the first r columns of AP are also linearly independent, NONE of the P(ej) for 1 ≤ j ≤ r, are in L, and the only linear combination of them

which is in L, is the 0-vector.

so {P(e(r+1)),...,P(en)} span L, and are thus a basis for L, which therefore has dimension n - r.