We are ready to prove that the fundamental solutions of $A\mathbf{x}=\mathrm{\U0001d7ce}$ are a basis for $N(A)$. We use the notation of Section 4.11 where we proved the fundamental solutions were linearly independent: $A$ is a $m\times n$ matrix, $R$ is a RREF matrix obtained by doing row operations to $A$, the number of columns of $R$ with a leading entry is $r$ and the number of columns with no leading entry is $k$, so $r+k=n$. There are $k$ fundamental solutions to $A\mathbf{x}=\mathrm{\U0001d7ce}$, and we showed in Lemma 4.11.2 that these are linearly independent.

The fundamental solutions to $A\mathit{}\mathbf{x}\mathrm{=}\mathrm{\U0001d7ce}$ are a basis of the nullspace $N\mathit{}\mathrm{(}A\mathrm{)}$.

Consider the linear map ${T}_{R}:{\mathbb{F}}^{n}\to {\mathbb{F}}^{m}$. The kernel of ${T}_{R}$, which is the nullspace $N(R)$, contains the $k$ fundamental solutions, which are linearly independent, so $dim\mathrm{ker}{T}_{R}\u2a7ek$ by Corollary 4.13.1.

The image of ${T}_{R}$, which is the column space $C(R)$, contains each of the $r$ columns of $R$ which contain a leading entry. These are standard basis vectors (by definition of RREF), so by Corollary 4.13.1 again $dim\mathrm{im}{T}_{R}\u2a7er$.

We know that $k+r=n$, and the rank-nullity theorem says that $dim\mathrm{ker}{T}_{R}+dim\mathrm{im}{T}_{R}=n$. So $dim\mathrm{ker}{T}_{R}=k$ and $dim\mathrm{im}{T}_{R}=r$ (if $dim\mathrm{ker}{T}_{R}$ were strictly larger than $k$, for example, then $dim\mathrm{ker}{T}_{R}+dim\mathrm{im}{T}_{R}$ would be strictly larger than $k+r=n$, a contradiction).