Thursday, May 10, 2012

Rank of a Matrix Continued.....

In previous sections, we solved linear systems using Gauss elimination method or the Gauss-Jordan method. In the examples considered, we have encountered three possibilities, namely
  1. existence of a unique solution,
  2. existence of an infinite number of solutions, and
  3. no solution.
Based on the above possibilities, we have the following definition.


DEFINITION 2.4.1 (Consistent, Inconsistent)   A linear system is called CONSISTENT if it admits a solution and is called INCONSISTENT if it admits no solution.
The question arises, as to whether there are conditions under which the linear system $ A {\mathbf x}= {\mathbf b}$ is consistent. The answer to this question is in the affirmative. To proceed further, we need a few definitions and remarks.
Recall that the row reduced echelon form of a matrix is unique and therefore, the number of non-zero rows is a unique number. Also, note that the number of non-zero rows in either the row reduced form or the row reduced echelon form of a matrix are same.


DEFINITION 2.4.2 (Row rank of a Matrix)   The number of non-zero rows in the row reduced form of a matrix is called the row-rank of the matrix.
By the very definition, it is clear that row-equivalent matrices have the same row-rank. For a matrix $ A,$ we write ` $ {\mbox{row-rank }}(A)$ ' to denote the row-rank of $ A.$

EXAMPLE 2.4.3  
  1. Determine the row-rank of $ A = \begin{bmatrix}1 &
2 & 1 \\ 2 & 3 & 1 \\ 1 & 1 & 2 \end{bmatrix}.$
    Solution: To determine the row-rank of $ A,$ we proceed as follows.
    1. $ \begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1
& 1 & 2 \end{bmatrix} \overrightarr...
..._{31}(-1)}
\begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & 1
\end{bmatrix}.$
    2. $ \begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & 1
\end{bmatrix} \overright...
..., R_{32}(1) }
\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2
\end{bmatrix}.$
    3. $ \begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 2
\end{bmatrix}\overrightarro...
...R_{12}(-2) }
\begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}.$
    4. $ \begin{bmatrix}1 & 0 & -1 \\ 0 & 1 & 1 \\ 0 & 0 & 1
\end{bmatrix}\overrightarr...
...1), R_{13}(1)}\begin{bmatrix}1 & 0 & 0
\\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $
    The last matrix in Step 1d is the row reduced form of $ A$ which has $ 3$ non-zero rows. Thus, $ {\mbox{row-rank}}(A)~=~3.$ This result can also be easily deduced from the last matrix in Step 1b.
  2. Determine the row-rank of $ A = \begin{bmatrix}1 & 2 & 1
\\ 2 & 3 & 1 \\ 1 & 1 & 0 \end{bmatrix}.$
    Solution: Here we have
    1. $ \begin{bmatrix}1 & 2 & 1 \\ 2 & 3 & 1 \\ 1
& 1 & 0 \end{bmatrix} \overrightarr...
...31}(-1) }
\begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & -1
\end{bmatrix}.$
    2. $ \begin{bmatrix}1 & 2 & 1 \\ 0 & -1 & -1 \\ 0 & -1 & -1
\end{bmatrix} \overrigh...
..., R_{32}(1) }
\begin{bmatrix}1 & 2 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0
\end{bmatrix}.$
    From the last matrix in Step 2b, we deduce $ {\mbox{row-rank}}(A)=2.$


Remark 2.4.4   Let $ A {\mathbf x}= {\mathbf b}$ be a linear system with $ m$ equations and $ n$ unknowns. Then the row-reduced echelon form of $ A$ agrees with the first $ n$ columns of $ [A \; \; {\mathbf b}],$ and hence
$\displaystyle {\mbox{row-rank}}
(A) \leq {\mbox{row-rank}} ([A \; \; {\mathbf b}]).$

The reader is advised to supply a proof.


Remark 2.4.5   Consider a matrix $ A.$ After application of a finite number of elementary column operations (see Definition 2.3.16) to the matrix $ A,$ we can have a matrix, say $ B,$ which has the following properties:
  1. The first nonzero entry in each column is $ 1.$
  2. A column containing only 0 's comes after all columns with at least one non-zero entry.
  3. The first non-zero entry (the leading term) in each non-zero column moves down in successive columns.
Therefore, we can define column-rank of $ A$ as the number of non-zero columns in $ B.$ It will be proved later that

$\displaystyle {\mbox{row-rank}} (A) = {\mbox{column-rank}} (A).$

Thus we are led to the following definition.

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