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


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9 Answers 9

up vote 26 down vote accepted


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What is n? Also, isn't it true that you can have things other than zeros and ones in the reduced echelon form? – user2820379 Mar 13 at 0:15
@user2820379 $n$ is the size of $A$. Also, as emphasized in the answer, $A$ is reduced to a matrix that is in both row echelon form and column echelon form. Therefore, everything off-diagonal would be zeroed out (by a row operation or a column operation) and what remain on the diagonal are ones and zeros. – user1551 Mar 13 at 22:40


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An alternative 'symmetric' definition for the rank might help here.

Define the rank of $A \in F^{m\times n} $ as the minimal $k$ for which $A$ can be represented as $$ A = c_1 r_1 + \cdots + c_k r_k$$ where $c_i \in F^{m\times 1}$ and $r_i \in F^{1 \times n}$.

Then it follows that each row of $A$ is a linear combination of $r_1, \cdots, r_k$ since the $i$th row of $A$ will be $(c_1)_i r_1 + \cdots + (c_k)_i r_k$. This implies $\text{row rank} \le k$

On the other hand, if we let $r'_1, \cdots, r'_t$ be a basis of the row space, we can write each of row of $A$ as a linear combination of $r'_i$s. Then by collecting the coefficients of $r'_i$ and making it a column vector $c'_i$, we have $A = c'_1 r'_1 + \cdots + c'_t r'_t$. Thus $\text{row rank} = t \ge k$ by the minimality of $k$.

Summing these up, we have $k = \text{row rank}$. And with the same arguments, once can easily prove $k = \text{column rank}$.

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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

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

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