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.
right multiplying a nonzero scalar to a row

6.
left multiplying a nonzero scalar to a row

7.
right multiplying a nonzero scalar to a column

8.
left multiplying a nonzero 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 ${v}_{i}=({a}_{i1},\mathrm{\cdots},{v}_{im})$ be the $i$th row of $M$, and ${w}_{i}=({b}_{i1},\mathrm{\cdots},{b}_{im})$ be the $i$th row of ${M}^{\prime}$. In other words,
$$M=\left(\begin{array}{c}\hfill {v}_{1}\hfill \\ \hfill \mathrm{\vdots}\hfill \\ \hfill {v}_{n}\hfill \end{array}\right)=\left(\begin{array}{ccc}\hfill {a}_{11}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {a}_{1m}\hfill \\ \hfill \mathrm{\vdots}\hfill & \hfill \mathrm{\ddots}\hfill & \hfill \mathrm{\vdots}\hfill \\ \hfill {a}_{n1}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {a}_{nm}\hfill \end{array}\right)\underset{\text{operation}}{\overset{\text{elementary}}{\to}}\left(\begin{array}{ccc}\hfill {b}_{11}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {b}_{1m}\hfill \\ \hfill \mathrm{\vdots}\hfill & \hfill \mathrm{\ddots}\hfill & \hfill \mathrm{\vdots}\hfill \\ \hfill {b}_{n1}\hfill & \hfill \mathrm{\cdots}\hfill & \hfill {b}_{nm}\hfill \end{array}\right)=\left(\begin{array}{c}\hfill {w}_{1}\hfill \\ \hfill \mathrm{\vdots}\hfill \\ \hfill {w}_{n}\hfill \end{array}\right)={M}^{\prime}$$ 
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 ${w}_{1},\mathrm{\dots},{w}_{n}$ are vectors obtained respectively from ${v}_{1},\mathrm{\dots},{v}_{n}$ by exchanging the $i$th and $j$th coordinates^{}. Suppose ${r}_{1}{w}_{1}+\mathrm{\cdots}+{r}_{d}{w}_{d}=0$. Then we get an equation ${r}_{1}{b}_{1k}+\mathrm{\cdots}+{r}_{d}{b}_{dk}=0$ for $1\le k\le m$. Rearranging these equations, we see that ${r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{d}{v}_{d}=0$, which implies ${r}_{1}=\mathrm{\cdots}={r}_{d}=0$, showing that ${w}_{1},\mathrm{\dots},{w}_{d}$ 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 ${w}_{1},\mathrm{\dots},{w}_{n}$ be vectors obtained respectively from ${v}_{1},\mathrm{\dots},{v}_{n}$ 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 ${v}_{1},\mathrm{\dots},{v}_{d}$ be left linearly independent. Suppose ${r}_{1}{w}_{1}+\mathrm{\cdots}+{r}_{i}{w}_{i}+\mathrm{\cdots}+{r}_{d}{w}_{d}=0$. A similar argument like in the previous proposition shows that ${r}_{1}{v}_{1}+\mathrm{\cdots}+({r}_{i}+{r}_{j}){v}_{j}+{r}_{d}{v}_{d}=0$, which implies ${r}_{1}=\mathrm{\cdots}={r}_{i}+{r}_{j}=\mathrm{\cdots}{r}_{d}=0$. Since ${r}_{i}=0$, ${r}_{j}=0$ too. This shows that ${w}_{1},\mathrm{\dots},{w}_{d}$ 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) nonzero row scalar multiplication preserves left (right) row rank of $M$; left (right) nonzero column scalar multiplications preserves left (right) column rank of $M$.
Proof.
Let ${w}_{1},\mathrm{\dots},{w}_{n}$ be vectors obtained respectively from ${v}_{1},\mathrm{\dots},{v}_{n}$ such that the $i$th vector ${w}_{i}=r{v}_{i}$, where $0\ne r\in D$, 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 $i\le d$. Suppose ${r}_{1}{w}_{1}+\mathrm{\cdots}+{r}_{d}{w}_{d}=0$. Then ${r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{i}(r{v}_{i})+\mathrm{\cdots}{r}_{d}{v}_{d}=0$, which implies ${r}_{1}=\mathrm{\cdots}={r}_{i}r=\mathrm{\cdots}={r}_{d}=0$. Since $r\ne 0$, ${r}_{i}=0$, and therefore ${w}_{1},\mathrm{\dots},{w}_{d}$ are left linearly independent.
The others are proved similarly. ∎
Proposition 4.
Left (right) nonzero row scalar multiplication preserves right (left) column rank of $M$; left (right) nonzero column scalar multiplication preserves right (left) row rank of $M$.
Proof.
Let us prove that right multiplying a column by a nonzero scalar $r$ preserves the left row rank $d$ of $M$. The others follow similarly.
Let ${w}_{1},\mathrm{\dots},{w}_{n}$ be vectors obtained respectively from ${v}_{1},\mathrm{\dots},{v}_{n}$ 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 ${r}_{1}{w}_{1}+\mathrm{\cdots}+{r}_{d}{w}_{d}=0$. Then for each coordinate $j$ we get an equation ${r}_{1}{b}_{1j}+\mathrm{\cdots}+{r}_{d}{b}_{dj}=0$. In particular, for the $i$th coordinate, we have ${r}_{1}{a}_{1j}r+\mathrm{\cdots}+{r}_{d}{a}_{dj}r=0$. Since $r\ne 0$, right multiplying the equation by ${r}^{1}$ gives us ${r}_{1}{a}_{1j}+\mathrm{\cdots}+{r}_{d}{a}_{dj}=0$. Recollecting all the equations, we get ${r}_{1}{v}_{1}+\mathrm{\cdots}+{r}_{d}{w}_{d}=0$, which implies that ${r}_{1}=\mathrm{\cdots}={r}_{d}=0$, or that ${w}_{1},\mathrm{\dots},{w}_{d}$ are left linearly independent. ∎
Title  elementary matrix operations as rank preserving operations 

Canonical name  ElementaryMatrixOperationsAsRankPreservingOperations 
Date of creation  20130322 19:22:57 
Last modified on  20130322 19:22:57 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  18 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 1501 
Classification  msc 15A33 
Classification  msc 15A03 