The proof that column rank = row rank for matrices over a field relies on the fact that the elements of a field commute. I'm looking for an easy example of a matrix over a ring for which column rank $\neq$ row rank. i.e. can one find a $2 \times 3$-(block)matrix with real $2\times 2$-matrices as elements, which has different column and row ranks?

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    $\begingroup$ How do you define these ranks? $\endgroup$ Nov 25, 2010 at 19:54
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    $\begingroup$ For matrices over skew field one usually defines a left column rank, a right column rank, a left row rank and a right row rank. Then one still has the equalities left column rank = right row rank and right column rank = left row rank. $\endgroup$
    – Someone
    Nov 26, 2010 at 8:43

3 Answers 3


Let $D$ be a skew field and consider the sets of $2\times 1$-matrices (columns) and $1\times 2$-matrices (lines) as left vector spaces over $D$. Let $a$ and $b$ be two non-commuting elements of $D$. Then $(a,ab)\in D(1,b)$, on the other hand $(b,ab)^{\rm T}\not\in D(1,a)^{\rm T}$.

In particular the matrix $$ \left(\begin{array}{cc} 1 & b\\ a & ab \end{array} \right) $$ is not invertible, but its transpose $$ \left(\begin{array}{cc} 1 & a\\ b & ab \end{array} \right) $$ is invertible.


It is a classical observation due to Nathan Jacobson that a division ring such that the set of invertible matrices over it is closed under transposition has to be a field, i.e. commutative.

The reason is simple: the matrix $\begin{pmatrix} a & b \\ c & 1 \end{pmatrix}$ is invertible if and only if $\begin{pmatrix} a - bc & 0 \\ c & 1 \end{pmatrix}$ is invertible. This happens if and only $a - bc \neq 0$.

For the transpose you get the condition $a - cb \neq 0$. Hence, taking $a = cb$ and a pair of non-commuting elements $b,c$ in the division ring, you get an example of an invertible matrix, whose transpose is not invertible.

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    $\begingroup$ I suppose the hypothesis that $D$ is a division ring is made so that $a-bc$ is a unit. $\endgroup$
    – Eric Auld
    Feb 25, 2015 at 7:37
  • $\begingroup$ Yes, this is correct. $\endgroup$ Feb 25, 2015 at 9:11

I will only consider matrices over division rings, since defining rank over more general rings, in particular those with zero divisors, involves difficulties rather independent of those posed by non-commutativity.

Contrary to what is implicit in the question, and was stated explicitly in Wikipedia,* and in seeming contradiction with the other answers given here (although what they say is correct), I maintain that there is only one natural notion of rank for a matrix over a division ring$~D$. That is, all natural definitions give the same notion of rank; in particular the row rank of a matrix always equals its column rank, if these notions are defined in a natural way. The point is that the inherent asymmetry of matrix multiplication makes that the set of column vectors naturally forms only a right $D$ vector space, while the set of row vectors forms a left $D$ vector space. Of course since the two sets are in bijection one can define an opposite vector space structure, but to do so in terms of matrix multiplication requires using an "unnatural" operation such as (1)$~$matrix transposition while (still) viewing the result as matrix over $D$, (2)$~$wrong matrix multiplication with the roles of rows and columns interchanged with respect to the usual definition, or (3)$~$reinterpreting a matrix over$~D$ as a matrix over the opposite division ring$~D^\mathrm{op}$ (the three transgressions can be seen to be more or less equivalent). For this reason it is natural to define the column rank of a matrix to be what some call the right column rank, and the row rank to be the left row rank (these are well known to be equal); the "switched" notions of left column rank and right row rank just work the wrong way, and would only be important in a setting where wrong matrix multiplication is treated on an equal footing with ordinary matrix multiplication, which fortunately nobody does.

So the rank of a $n\times m$ matrix $A$ over$~D$ can be defined in any one of the following ways, all giving the same value

  • The dimension of the image of the linear map$~f$ of right $D$ vector spaces with matrix$~A$ (with respect to some bases); in coordinates, $f$ is given by left multiplication by$~A$ on column vectors.

  • The dimension of the image of the linear map$~f$ of left $D$ vector spaces with matrix$~A$ (with respect to some bases); in coordinates, $f$ is given by right multiplication by$~A$ on row vectors.

  • The dimension of the column space of $A$, the subspace of the right $D$ vector space $D^n$ spanned by the columns of $A$.

  • The dimension of the row space of $A$, the subspace of the left $D$ vector space $D^m$ spanned by the rows of $A$.

  • The decomposition rank of $A$: the smallest number$~r$ such that one can write $A=BC$ for some $n\times r$ matrix $B$ and $r\times m$ matrix $C$.

  • The largest number$~r$ such that there exist matrices $P,Q$ for which $PAQ=I_r$, the $r\times r$ identity matrix.

*Until I modified the article, which was after I posted this answer.

I'll add a word about the seeming contradiction which the other two answers. Rather than answering the stated question about row and columns ranks, they provide an example for the situation where the rank of the transpose of $A$ differs from that of$~A$. But they are implicitly interpreting $A^T$ as a matrix over$~D$, which is the first of the unnatural things I listed above. It is unnatural because one wants transposition the have some relation with composition; indeed one wants $(AB)^T=B^TA^T$ to hold, and this is only possible if $A^T$ is considered to be a matrix over$~D^\mathrm{op}$; doing so, transposition is a natural operation. But with that interpretation, the examples disappear, and the rank of $A^T$ is equal to the rank of$~A$.


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