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Homerank of a matrix

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# 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 $\operatorname{rc.rnk}(M)$, $\operatorname{lc.rnk}(M)$, $\operatorname{rr.rnk}(M)$, and $\operatorname{lr.rnk}(M)$.

Since the columns of $M$ are the rows of its transpose $M^{T}$, we have

$\operatorname{lc.rnk}(M)=\operatorname{lr.rnk}(M^{T}),\qquad\textrm{and}\qquad% \operatorname{rc.rnk}(M)=\operatorname{rr.rnk}(M^{T}).$ |

In addition, it can be shown that for a given matrix $M$,

$\operatorname{lc.rnk}(M)=\operatorname{rr.rnk}(M),\qquad\textrm{and}\qquad% \operatorname{rc.rnk}(M)=\operatorname{lr.rnk}(M).$ |

For any $0\neq r\in D$, 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, $\operatorname{lc.rnk}(M)=\operatorname{rc.rnk}(M)$, so that all four numbers are the same, and we simply call this number the *rank* of $M$, denoted by $\operatorname{rank}(M)$.

Rank can also be defined for matrices $M$ (over a fixed $D$) that satisfy the identity $M=rM^{T}$, 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 $\mathbb{H}$. They are columns in the $2\times 2$ matrix

$M:=\begin{pmatrix}1&i\\ j&k\end{pmatrix}$ |

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 $2$.

## Mathematics Subject Classification

15A03*no label found*15A33

*no label found*

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