# Intuitive proof of row rank = column rank? [duplicate]

Is it possible to give an intuitive/elementary proof of the theorem that says that the row rank of a (finite-dimensional) square matrix matrix equals its column rank?

• Here are two: en.wikipedia.org/wiki/… May 6, 2013 at 13:44
• The right understanding, I think, is what's in the second proof at the link @vadim123 provided. The point is that multiplying by the matrix $A$ (applying the linear map) gives a one-to-one correspondence between the row space of $A$ and the column space of $A$. May 6, 2013 at 14:42

Let $$\mathbb F$$ be a field and let $$T: \mathbb F^n \to \mathbb F^m$$ be a linear transformation. Then there exists an $$m$$ by $$n$$ matrix $$A$$ such that $$T(x)=Ax.$$ Let $$A_i \text{ (where }i = 1,...,n)$$ be the $$i$$th column of $$A$$. Then
$$T(x)=A_1x_1+A_2x_2+....+A_nx_n$$, so that $$\text{rank}(T) =$$ dimension of the subspace spanned by the columns of $$A$$. On the other hand,$$\text{rank}(T)+\text{null}(T)=n$$ since $$T$$ is a linear transformation from $$\mathbb F^n$$ to $$\mathbb F^m$$.
By definition, the $$\text{null}(T)$$ is the dimension of the nullspace of $$T$$, where the nullspace of $$T$$ is $$\{x \in \mathbb F^n : Ax=0\}$$. Thus $$\text{null}(T)= n- \text{dimension of row space of A}$$. By rank nullity theorem, we thus have $$\text{column space of A} + n - \text{row space of }A = n$$. Thus, dimension of the column space of $$A$$ is equal to the dimension of the row space of $$A$$.
• It's unclear about what do you mean $null(T)$? null space or the nullity? Sep 20, 2020 at 12:07
• You define $rank(T)$ as rank of column space of $A$ and you said $\dim(null(T))=n-\textrm{rank of row space of }A$, which implicitly use "row rank=column rank". Sep 20, 2020 at 12:46