The Magic of Exact Sequences: Unveiling the Equivalence of Matrix Ranks

 

The aim of this article is to prove that $\mathrm{Rank}(A)=\mathrm{Rank}(A^T)$$\mathrm{Rank}(A)=\mathrm{Rank}(A^T)$ elegantly.

Consider the following exact sequence in ${\mathrm{FinVect}}_{\mathbb{k}}$$\mathrm{FinVect}_{\mathbb k}$.

$$\begin{array}{}\text{(1)}& 0\stackrel{}{\to }\mathrm{Ker}(f)\stackrel{i}{\to }V\stackrel{f}{\to }W\stackrel{\pi }{\to }\mathrm{Coker}(f)\stackrel{}{\to }0\end{array}$$

Then applied the dual space functor $\mathrm{Hom}(-,\mathbb{k})$$\mathrm{Hom}(-,\mathbb k)$, which is exact.

Hence we get a new exact sequence

$$\begin{array}{}\text{(2)}& 0\stackrel{}{\to }\mathrm{Coker}(f{)}^{\ast }\stackrel{{\pi }^{\ast }}{\to }{W}^{\ast }\stackrel{f^{\ast }}{\to }{V}^{\ast }\stackrel{i^{\ast }}{\to }\mathrm{Ker}(f{)}^{\ast }\stackrel{}{\to }0\end{array}$$

Hence we get that

$$\begin{array}{}\text{(3)}& \mathrm{Coker}(f{)}^{\ast }\cong \mathrm{Ker}(f^{\ast })\end{array}$$

$$\begin{array}{}\text{(4)}& \dim \mathrm{Im}{f}^{\ast }=\dim W^{\ast }-\dim \mathrm{Ker}(f^{\ast })=\dim W-\dim \mathrm{Coker}(f{)}^{\ast }=\dim W-\dim \mathrm{Coker}(f)\end{array}$$

But

$$\begin{array}{}\text{(5)}& \dim W-\dim \mathrm{Coker}(f)=\dim W-(\dim W-\dim \mathrm{Im}f)=\dim \mathrm{Im}(f)\end{array}$$

Hence we get that

$$\begin{array}{}\text{(6)}& \dim \mathrm{Im}(f)=\dim \mathrm{Im}(f^{\ast })\end{array}$$