Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
How do you define these ranks? –  darij grinberg Nov 25 '10 at 19:54
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. –  Someone Nov 26 '10 at 8:43
add comment

2 Answers

up vote 18 down vote accepted

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.

share|improve this answer
add comment

It is a classical observation due to Nathan Jacobson that a division ring such the set of invertible matrices 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.

share|improve this answer
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.