# elementary matrix operations as rank preserving operations

Let $M$ be a matrix over a division ring $D$. An elementary operation on $M$ is any one of the eight operations below:

1. 1.

exchanging two rows

2. 2.

exchanging two columns

3. 3.

4. 4.

5. 5.

right multiplying a non-zero scalar to a row

6. 6.

left multiplying a non-zero scalar to a row

7. 7.

right multiplying a non-zero scalar to a column

8. 8.

left multiplying a non-zero scalar to a column

We want to determine the effects of these operations on the various ranks of $M$. To facilitate this discussion, let $M=(a_{ij})$ be an $n\times m$ matrix and $M^{\prime}=(b_{ij})$ be the matrix after an application of one of the operations above to $M$. In addition, let $v_{i}=(a_{i1},\cdots,v_{im})$ be the $i$-th row of $M$, and $w_{i}=(b_{i1},\cdots,b_{im})$ be the $i$-th row of $M^{\prime}$. In other words,

 $M=\begin{pmatrix}v_{1}\\ \vdots\\ v_{n}\end{pmatrix}=\begin{pmatrix}a_{11}&\cdots&a_{1m}\\ \vdots&\ddots&\vdots\\ a_{n1}&\cdots&a_{nm}\end{pmatrix}\xrightarrow[\textrm{operation}]{\textrm{% elementary}}\begin{pmatrix}b_{11}&\cdots&b_{1m}\\ \vdots&\ddots&\vdots\\ b_{n1}&\cdots&b_{nm}\end{pmatrix}=\begin{pmatrix}w_{1}\\ \vdots\\ w_{n}\end{pmatrix}=M^{\prime}$

Finally, let $d$ be the left row rank of $M$.

###### Proposition 1.

Row and column exchanges preserve all ranks of $M$.

###### Proof.

Clearly, exchanging two rows of $M$ do not change the subspace generated by the rows of $M$, and therefore $d$ is preserved.

As exchanging rows do not affect $d$, let us assume that rows have been exchanged so that the first $d$ rows of $M$ are left linearly independent.

Now, let $M^{\prime}$ be obtained from $M$ by exchanging columns $i$ and $j$. So $w_{1},\ldots,w_{n}$ are vectors obtained respectively from $v_{1},\ldots,v_{n}$ by exchanging the $i$-th and $j$-th coordinates. Suppose $r_{1}w_{1}+\cdots+r_{d}w_{d}=0$. Then we get an equation $r_{1}b_{1k}+\cdots+r_{d}b_{dk}=0$ for $1\leq k\leq m$. Rearranging these equations, we see that $r_{1}v_{1}+\cdots+r_{d}v_{d}=0$, which implies $r_{1}=\cdots=r_{d}=0$, showing that $w_{1},\ldots,w_{d}$ are left linearly independent. This means that $d$ is preserved by column exchanges.

Preservation of other ranks of $M$ are similarly proved. ∎

###### Proposition 2.

Additions of rows and columns preserve all ranks of $M$.

###### Proof.

Let $M^{\prime}$ be the matrix obtained from $M$ by replacing row $i$ by vector $v_{i}+v_{j}$, and let $V^{\prime}$ be the left vector space spanned by the rows of $M^{\prime}$. Since $v_{i}+v_{j}\in V$, we have $V^{\prime}\subseteq V$. On other hand, $v_{i}=(v_{i}+v_{j})-v_{j}\in V^{\prime}$, so $V\subseteq V^{\prime}$, and hence $V=V^{\prime}$.

Next, let $w_{1},\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\ldots,v_{n}$ such that the $i$-th coordinate of $w_{k}$ is the sum of the $i$-th coordinate of $v_{k}$ and the $j$-th coordinate of $v_{k}$, with all other coordinates remain the same. Again, by renumbering if necessary, let $v_{1},\ldots,v_{d}$ be left linearly independent. Suppose $r_{1}w_{1}+\cdots+r_{i}w_{i}+\cdots+r_{d}w_{d}=0$. A similar argument like in the previous proposition shows that $r_{1}v_{1}+\cdots+(r_{i}+r_{j})v_{j}+r_{d}v_{d}=0$, which implies $r_{1}=\cdots=r_{i}+r_{j}=\cdots r_{d}=0$. Since $r_{i}=0$, $r_{j}=0$ too. This shows that $w_{1},\ldots,w_{d}$ are left linearly independent, which means that $d$ is preserved by additions of columns.

Preservation of other ranks of $M$ are proved similarly. ∎

###### Proposition 3.

Left (right) non-zero row scalar multiplication preserves left (right) row rank of $M$; left (right) non-zero column scalar multiplications preserves left (right) column rank of $M$.

###### Proof.

Let $w_{1},\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\ldots,v_{n}$ such that the $i$-th vector $w_{i}=rv_{i}$, where $0\neq r\in D$, and all other $w_{j}$’s are the same as the $v_{j}$’s. Assume that the first $d$ rows of $M$ are left linearly independent, and that $i\leq d$. Suppose $r_{1}w_{1}+\cdots+r_{d}w_{d}=0$. Then $r_{1}v_{1}+\cdots+r_{i}(rv_{i})+\cdots r_{d}v_{d}=0$, which implies $r_{1}=\cdots=r_{i}r=\cdots=r_{d}=0$. Since $r\neq 0$, $r_{i}=0$, and therefore $w_{1},\ldots,w_{d}$ are left linearly independent.

The others are proved similarly. ∎

###### Proposition 4.

Left (right) non-zero row scalar multiplication preserves right (left) column rank of $M$; left (right) non-zero column scalar multiplication preserves right (left) row rank of $M$.

###### Proof.

Let us prove that right multiplying a column by a non-zero scalar $r$ preserves the left row rank $d$ of $M$. The others follow similarly.

Let $w_{1},\ldots,w_{n}$ be vectors obtained respectively from $v_{1},\ldots,v_{n}$ such that the $i$-th coordinate $b_{ik}$ of $w_{k}$ is $a_{ik}r$, where $a_{ik}$ is the $i$-th coordinate of $v_{k}$. Suppose once again that the first $d$ rows of $M$ are left linearly independent, and suppose $r_{1}w_{1}+\cdots+r_{d}w_{d}=0$. Then for each coordinate $j$ we get an equation $r_{1}b_{1j}+\cdots+r_{d}b_{dj}=0$. In particular, for the $i$-th coordinate, we have $r_{1}a_{1j}r+\cdots+r_{d}a_{dj}r=0$. Since $r\neq 0$, right multiplying the equation by $r^{-1}$ gives us $r_{1}a_{1j}+\cdots+r_{d}a_{dj}=0$. Re-collecting all the equations, we get $r_{1}v_{1}+\cdots+r_{d}w_{d}=0$, which implies that $r_{1}=\cdots=r_{d}=0$, or that $w_{1},\ldots,w_{d}$ are left linearly independent. ∎

Title elementary matrix operations as rank preserving operations ElementaryMatrixOperationsAsRankPreservingOperations 2013-03-22 19:22:57 2013-03-22 19:22:57 CWoo (3771) CWoo (3771) 18 CWoo (3771) Definition msc 15-01 msc 15A33 msc 15A03