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 non-zero scalar to a row
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. ∎

Title	elementary matrix operations as rank preserving operations
Canonical name	ElementaryMatrixOperationsAsRankPreservingOperations
Date of creation	2013-03-22 19:22:57
Last modified on	2013-03-22 19:22:57
Owner	CWoo (3771)
Last modified by	CWoo (3771)
Numerical id	18
Author	CWoo (3771)
Entry type	Definition
Classification	msc 15-01
Classification	msc 15A33
Classification	msc 15A03