The Rank Nullity Theorem

Theorem.

If $T\in \mathcal{L}(V,W)$, then $dim(V)$ is the sum of $dim(null(T))$ and $dim(range(T))$.

Proof.

Suppose $X=null(T)$. Since every subspace of V has a complement, there exists a vector space $Y\subseteq V$ such that $V=X⨁Y$. The directness of the sum implies that $dim(V)=dim(X)+dim(Y)$. Thus, we need only show that $dim(Y)=dim(range(T))$. To this end, we conjure a linear map $U:Y\to range(T),u\mapsto Tu$ which we will argue is bijective. Injectivity is shown by the fact that $Y\cap null(T)=\{0\}$ since $Y$ is the complement of $null(T)$. For surjectivity, suppose that $Tu\ne 0$. Since $V=X⨁Y$, $u$ can be represented as a unique sum of some $x\in X$ and some $y\in Y$, i.e.

$$Tu=T(x+y)=Tx+Ty=0+Ty⟹u=y\in Y,$$

where we used the fact that $T$ is injective. Thus, $U(y)=Tu$ for some $y$ and $U$ is surjective. $\square$

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A nice thing about this proof is that we never had to invoke any bases. We simply asserted the existence of a complement of $X$ without ever constructing it. The crux of this argument is of course the existence of complements. Zorn’s Lemma implies the existence of complements in the infinite-dimensional case, so this proof scales nicely.