Looking for an intuitive explanation why the row rank is equal to the column rank for a matrix

I am looking for an intuitive explanation as to why/how row rank of a matrix = column rank. I've read the proof on Wikipedia and I understand the proof, but I don't "get it". Can someone help me out with this ?

I find it hard to wrap my head around the idea of how the column space and the row space is related at a fundamental level.

Solution 1:

You can apply elementary row operations and elementary column operations to bring a matrix $A$ to a matrix that is in both row reduced echelon form and column reduced echelon form. In other words, there exist invertible matrices $P$ and $Q$ (which are products of elementary matrices) such that $$PAQ=E:=\begin{pmatrix}I_k\\&0_{(n-k)\times(n-k)}\end{pmatrix}.$$ As $P$ and $Q$ are invertible, the maximum number of linearly independent rows in $A$ is equal to the maximum number of linearly independent rows in $E$. That is, the row rank of $A$ is equal to the row rank of $E$. Similarly for the column ranks. Now it is evident that the row rank and column rank of $E$ are identical (to $k$). Hence the same holds for $A$.

Solution 2:

This post is quite old, so my answer might come a bit late. If you are looking for an intuition (you want to "get it") rather than a demonstration (of which there are several), then here is my 5c.

If you think of a matrix A in the context of solving a system of simultaneous equations, then the row-rank of the matrix is the number of independent equations, and the column-rank of the matrix is the number of independent parameters that you can estimate from the equation. That I think makes it a bit easier to see why they should be equal.

Saad.

Solution 3:

Let $V,W$ be two vector spaces such that $\dim(V)=n$

The column rank of a matrix $A=\alpha_{(V,W)}(\varphi)$ is equal to the dimension of the image of the application $\varphi$, so

$$\text{column rank}=\dim(\text{im}(\varphi))$$

On the other side, the row rank has to do with the dimension of the kernel of $\varphi$, namely $$\dim(\ker(\varphi))=n-\text{row rank}$$ Now, we know from rank-nullity theorem that

$$\dim(\ker(\varphi))+\dim(\text{im}(\varphi))=\dim(V)=n$$ $$\implies n-\text{row rank}+\text{column rank}=n$$ $$\implies \text{column rank}=\text{row rank}$$