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 anti-symmetric 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 $ℍ$. They are columns in the $2\times 2$ matrix

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 $r,s\in ℍ$. 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 $2$.