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# 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. exchanging two rows

2. exchanging two columns

3. adding one row to another

4. adding one column to another

5. 6. left multiplying a non-zero scalar to a row

7. right multiplying a non-zero scalar to a column

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. ∎

## Mathematics Subject Classification

15-01*no label found*15A33

*no label found*15A03

*no label found*

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