Why is the dimension of the nullspace of A equal to the nullity of A?

Question

For a $m\times n$ matrix $A$, it seems that wikipedia DEFINES nullity(A)=dim(nullspace(A)), whereas my textbook covers the topic the way round: nullity(A) is defined to be n-rank(A). Can one prove that dim(nullspace(A))=nullity(A) for from this definition of nullity?

Answer 1

Proving your question is equal to proving the rank-nullity theorem, which is as follows:

Let $V$ and $W$ be two finite-dimensional vector spaces over a field $\mathbb{F}$. Then if $A:V\rightarrow W$ is any linear map, $\dim\ker A+\dim\mathrm{ran}\ A=\dim V$, where $\ker A$ is the nullspace of $A$ and $\mathrm{ran}\ A$ is the range of $A$. By definition $\mathrm{rank}\ A=\dim\mathrm{ran}\ A$.

Proof:

Let $E=\{e_1,...,e_n\}\subset V$ be a basis in $V$ (then $\dim V=n$), in such a way that $\{e_1,...,e_k\}$ is a basis for $\ker A$.

Then obviously $\dim\ker A=k$, and $Ae_1,...,Ae_k=0_W$, while $Ae_{k+1},...,Ae_n$ are linearly independent in $W$, since if they weren't linearly independent, there would be $\alpha_{k+1},...,\alpha_{n}\in\mathbb{F}$ scalars not all zero satisfying $$ 0_W=\alpha_{k+1}Ae_{k+1}+...+\alpha_nAe_n=\\=A(\alpha_{k+1}e_{k+1}+...+\alpha_ne_n), $$ which would either imply that $\{e_{k+1},...,e_{n}\}$ is linearly dependent, or that their span is a subspace of $\ker A$, obviously neither is true.

But the set $$ \{Ae_1,...,Ae_n\} $$ surely spans $\mathrm{ran}\ A$, since the set of all linear combination of the above set corresponds to every possible way of inputing vectors into $A$, and the first $k$ elements of the above set are zero, the remaining $n-k$ members span $\mathrm{ran}\ A$ and are linearly independent, therefore is a basis of $\mathrm{ran}\ A$.

We have proven that the dimensionality of the space $V$ is the sum of the dimensionality of the null space and range of any linear operator. From this, your question is proven as well.