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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 ?

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.

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This is one of the things that I have never "grokked" neither. It looks like a miracle. – Giuseppe Negro Mar 17 '13 at 16:50
Maybe Jordan's canonical form might be of some help. According to that theory, up to similitude every matrix $M$ can be written as $$M=D+N, $$where $D$ is diagonal and $N$ is nilpotent. Transposing, we get $$M^T=D+N^T, $$ so the matrix and its transpose have the same diagonal part. In particular $M$ and $M^T$ share any information that is carried by the diagonal part, and the (column) rank is one of them. This looks like excessively complicated but I cannot think of any simpler explanation. – Giuseppe Negro Mar 17 '13 at 16:55

12 Answers 12

up vote 28 down vote accepted

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 $A$ are 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 $A$.

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+1 Nice explanation. This essential point of this argument is that elementary row operations, which by construction don't alter the row rank, also no not alter the column rank (and similarly for column operations). That this is so is because doing an elementary row operation just amounts to expressing all columns in a different basis. – Marc van Leeuwen Mar 18 '13 at 5:40
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 '15 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 '15 at 22:40

Define the rank of a matrix $A$ as the largest size of any square submatrix (minor) with non-null determinant. Then if you see the columns of $A$ as vectors, the rank of $A$ can be thought of as the maximal number of linearly independent such vectors. Finally, note that $det(M)=det(M^T)$, and if $M$ is a minor of $A$, then $M^T$ is a minor of $A^T$.

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One way to view the rank $r$ of a $n\times m$ matrix $A$ with entries in a field $K$, is that it is the smallest number such that one can factor the linear map $f_A:K^m\to K^n$ corresponding to $A$ through an intermediate space of dimension$~r$, in other words, as a composition $K^m\to K^r\to K^n$ (taking $C$ and $B$ as the matrices corresponding to the two steps, this means that one has a decomposition $A=BC$ of $A$ as the product of a $n\times r$ and a $r\times m$ matrix). Now one can always factor $f_A$ though the image $\operatorname{Im}f_A\subseteq K^n$, as $K^m\to\operatorname{Im}f_A\hookrightarrow K^n$, and on the other hand this image can never have a dimension larger than a space through which $f_A$ factors; therefore the rank is equal to $\dim\operatorname{Im}f_A$. But that dimension is equal to the maximal number of independent columns of $A$, its column rank.

On the other hand one can view the rows of $A$ as linear functions on $K^m$ that describe the coordinates of $f_A(x)$ as a function of $x$, and the row rank $s$ is the maximum number of independent such functions; one such an independent set of $s$ independent rows is chosen, the remaining coordinates of $f_A(x)$ can each be described by a fixed linear combination of the chosen coordinates (because their rows are such linear combinations of the chosen rows). But this means that one can factor $f_A$ through $K^s$, with the map $K^s\to K^n$ reconstructing the dependent coordinates. The chosen coordinates are independent, so there is no nontrivial relation between them, and the map $K^m\to K^s$ is therefore surjective. This means that $f_A$ cannot factor through a space of smaller dimension than $s$, so the row rank $s$ is also equal to the rank of $A$.

Instead of that separate argument involving the row rank, you can also interpret the row rank of $A$ as the column rank of the transpose matrix $A^t$. Now one can factor $A=BC$ if and only if one can factor $A^t=C^tB^t$; then the minimal $r$ such that one write $A=BC$ with $B\in M_{n,r}$ and $C\in M_{r,m}$ (the column rank of $A$) obviously equals the minimal $r$ such that one write $A^t=C^tB^t$ with $C^t\in M_{m,r}$ and $B^t\in M_{r,n}$ (the column rank of $A^t$, and row rank of $A$).

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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.


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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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I think Strang's "four subspaces" picture is enlightening here. Assume $A \in \mathbb R^{m \times n}$. It's easy to prove that the null space of $A$ and the image of $A^T$ are orthogonal complements. Also, $A$ (as a mapping) is one to one when restricted to the range of $A^T$. So the range of $A^T$ is actually isomorphic to the range of $A$, and $A$ itself provides the isomorphism!

Strang presents the four subspace picture in the context of inner product spaces (in fact just $\mathbb R^n$). But the picture works for arbitrary finite dimensional vector spaces if we use annihilators instead of orthogonal complements. This gives an easy, conceptual proof that $A$ and $A^T$ have the same rank. See chapter two of Lax's linear algebra book for the (easy) details.

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Let $V$ be a vector space of dimension n, W of dimension $m$. If $M$ is m by n matrix, it gives a linear map from $V$ to $W$. The image of this map is the column space. The dual map $W^* \rightarrow V^*$ is given by the transpose $M^T$. The image of this map is therefore the row space. Further, the Hom functor is exact in this context, so the exact sequence: $$0 \rightarrow ker(M) \rightarrow V \overset M \rightarrow M(V) \rightarrow 0$$ gives an exact sequence:

$$ 0 \rightarrow M(V)^* \overset {M^T|_{M(V)^*}} \rightarrow V^* \rightarrow ker(M)^* \rightarrow 0$$

M(V) has the same dimension as $M(V)^*$. By exactness of Hom again, the image of the map $M^T|_{M(V)^*}$ (which is $M(V)^*$) coincides with the image of $M^T$, the row space.

Note: I think only left exactness of Hom is used here...

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Maybe this helps a bit with the intuition: When you transpose a matrix you don't change the dimension of the image. But when you transpose a matrix, the column rank becomes the row rank and vice versa. As the dimension of the image is the column rank those are equal.

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Saying that when you transpose a matrix you don't change the dimension of the image is just saying that the transpose has the same column rank as the original. How is this easier to see that that row and column rank are the same? – Marc van Leeuwen Sep 16 '13 at 7:22
@MarcvanLeeuwen well sometime just rephrasing a statement makes it easier to see – Dominic Michaelis Sep 16 '13 at 9:16
Agreed, but I don't see how that would apply to the current case. – Marc van Leeuwen Sep 16 '13 at 9:22

Given a linear system of equations:$$\begin{cases}ax+by = c_1\\ cx+dy = c_2\end{cases}$$ Is there a solution?

One way to answer this is; if the lines are not parallel, then there is a solution. From the slope-intercept form of the equation of a line: $y=mx+b$, $y=-\dfrac abx + \dfrac{c_1}b$ and $y=-\dfrac cdx + \dfrac{c_2}b$.

If the lines are to intersect, the slopes can not be equal: that is $\dfrac ab \neq \dfrac cd$, which gives: $ad \neq bc\implies ad-bc \neq0$ ....the determinant can not equal zero.

Alternatively, solving for $x$ instead: $x=-\dfrac ba + \dfrac{c_1}a$ and $x =-\dfrac dc + \dfrac{c_2}c$.

$\dfrac ba \neq \dfrac dc$ and again: $da-bc \neq 0$

You can also see that the coefficients of $x$ and $y$ are "intimately" related.

I doubt the above is rigourous enough for most people, but this helps me see the relationship between rows and columns $\ldots$ between coefficients.

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To see this without calculation, I guess the key is to realize that the dual basis and the basis of a matrix are one to one relationship. Therefore the ranks are the same.

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This can also be understood in terms of the singular value decomposition. Although proving the SVD takes a bit of work, once proved it provides a more thorough intuition about what is going on.

Geometrically, the SVD says that a matrix maps the unit (hyper-)sphere in the domain to a (hyper-)ellipsoid in the range, possibly squashing some axes of the ellipsoid to be flat. Moreover, the axes of the ellipsoid in the range correspond to an orthonormal set of vectors on the sphere in the domain.

The non-flat axes of the ellipsoid span the column space, and their corresponding axes on the sphere span the row space. Since these axes are orthonormal and in one-to-one correspondence, the fact that their dimension is the same becomes trivial.

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