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? 



Example for column rank $\neq$ row rank
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.


