rank of a matrix
Let $D$ be a division ring, and $M$ an $m\times n$ matrix over $D$. There are four numbers we can associate with $M$:

1.
the dimension^{} of the subspace^{} spanned by the columns of $M$ viewed as elements of the $n$dimensional right vector space^{} over $D$.

2.
the dimension of the subspace spanned by the columns of $M$ viewed as elements of the $n$dimensional left vector space over $D$.

3.
the dimension of the subspace spanned by the rows of $M$ viewed as elements of the $m$dimensional right vector space over $D$.

4.
the dimension of the subspace spanned by the rows of $M$ viewed as elements of the $m$dimensional left vector space over $D$.
The numbers are respectively called the right column rank, left column rank, right row rank, and left row rank of $M$, and they are respectively denoted by $\mathrm{rc}.\mathrm{rnk}(M)$, $\mathrm{lc}.\mathrm{rnk}(M)$, $\mathrm{rr}.\mathrm{rnk}(M)$, and $\mathrm{lr}.\mathrm{rnk}(M)$.
Since the columns of $M$ are the rows of its transpose^{} ${M}^{T}$, we have
$$\mathrm{lc}.\mathrm{rnk}(M)=\mathrm{lr}.\mathrm{rnk}({M}^{T}),\text{and}\mathit{\hspace{1em}\hspace{1em}}\mathrm{rc}.\mathrm{rnk}(M)=\mathrm{rr}.\mathrm{rnk}({M}^{T}).$$ 
In addition, it can be shown that for a given matrix $M$,
$$\mathrm{lc}.\mathrm{rnk}(M)=\mathrm{rr}.\mathrm{rnk}(M),\text{and}\mathit{\hspace{1em}\hspace{1em}}\mathrm{rc}.\mathrm{rnk}(M)=\mathrm{lr}.\mathrm{rnk}(M).$$ 
For any $0\ne r\in D$, it is also easy to see that the left column and row ranks of $rM$ are the same as those of $M$. Similarly, the right column and row ranks of $Mr$ are the same as those of $M$.
If $D$ is a field, $\mathrm{lc}.\mathrm{rnk}(M)=\mathrm{rc}.\mathrm{rnk}(M)$, so that all four numbers are the same, and we simply call this number the rank of $M$, denoted by $\mathrm{rank}(M)$.
Rank can also be defined for matrices $M$ (over a fixed $D$) that satisfy the identity^{} $M=r{M}^{T}$, where $r$ is in the center of $D$. Matrices satisfying the identity include symmetric^{} and antisymmetric matrices.
However, the left column rank is not necessarily the same as the right row rank of a matrix, if the underlying division ring is not commutative^{}, as can be shown in the following example: let $u=(1,j)$ and $v=(i,k)$ be vectors over the Hamiltonian quaternions $\mathbb{H}$. They are columns in the $2\times 2$ matrix
$$M:=\left(\begin{array}{cc}\hfill 1\hfill & \hfill i\hfill \\ \hfill j\hfill & \hfill k\hfill \end{array}\right)$$ 
Since $iu=(i,ij)=(i,k)=v$, they are left linearly dependent, and therefore the left column rank of $M$ is $1$. Now, suppose $ur+vs=(0,0)$, with $r,s\in \mathbb{H}$. Since $ui=(i,ji)=(i,k)$, then $ui(ir)+vs=0$, which boils down to two equations $ir=s$ and $ir=s$, and which imply that $s=r=0$, showing that $u,v$ are right linearly independent. Thus the right column rank of $M$ is $2$.
Title  rank of a matrix 
Canonical name  RankOfAMatrix 
Date of creation  20130322 19:22:42 
Last modified on  20130322 19:22:42 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  15 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 15A03 
Classification  msc 15A33 
Related topic  DeterminingRankOfMatrix 
Defines  left row rank 
Defines  left column rank 
Defines  right row rank 
Defines  right column rank 