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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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How do you define these ranks? –  darij grinberg Nov 25 '10 at 19:54
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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
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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.

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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.

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