Relationship Between Rank, Nullity, Determinant and Invertibility

Using the same notation in the theorem above, we prove this using the following lemmas.

Proof

Assume that $\mathrm{nullity}(T)=0$ and let $\mathit{a},\mathit{b}\in V$ such that $T(\mathit{a})=T(\mathit{b})$. This implies that

Then since $\mathrm{nullity}(T)=0$, the set of values $\mathit{a}-\mathit{b}$ which map to zero contains only the zero dimensional subspace $\{\mathbf{0}\}$, and hence $\mathit{a}-\mathit{b}=\mathbf{0}$ which implies $\mathit{a}=\mathit{b}$.

On the converse, if $T$ is assumed to be injective then

which is equivalent to

With $\mathit{a}-\mathit{b}$ an arbitrary vector by setting $\mathit{b}=0$, we have that the only thing that maps to zero is the zero vector itself, and therefore the subspace of elements mapped to zero is zero dimensional.

Proof

If $\mathrm{rank}(T)=\dim W$ then $T$ must be surjective, as an $n$ dimensional subspace of an $n$ dimensional space must be the whole space.

Similarly if $T$ is surjective, then $T(\mathit{x})$ is any arbitrary element of $W$, and hence $T$ is full rank.

Proof

We can now prove the main result. (2) and (3) are equivalent from the rank nullity theorem. Since (2) implies injectivity, and (3) implies surjectivity, but (2) is equivalent to (3), each of these independently prove bijectivity, (1). The converse is also true, with (1) implying $T$ is both surjective and injective, hence proving both (2) and (3).

Hence we have established $(1)⟺(2)⟺(3)$.

Then from matrix has non-zero determinant if and only if it is invertible we can conclude the final equivalenc.