Question

What is a rank \[1\] matrix?

Answer 1

Hint: Let us first know about a matrix. A matrix of order \[m \times n\] is a matrix in which we have \[m\] rows and \[n\] columns. In a matrix, if rows and columns are interchanged, then that matrix is called a transposed matrix. Consider a matrix \[A\] of order \[m \times n\] , and after interchanging the rows and columns, we get a transposed matrix denoted by \[{A^T}\] of order \[n \times m\] .

Complete solution:
The rank of a matrix is defined as the number of non-zero rows in that matrix, and is denoted by \[r(A)\] .
It can also be defined as, maximum number of linearly independent columns of that matrix.
If a matrix can be written as a non zero product of a row vector and a column vector, then it is said to be a “rank \[1\] matrix”.
Let \[A\] and \[B\] be two column vectors. Then \[P\] is said to be a matrix of rank \[1\] , where, \[P = A{B^T}\] .
Let us take an example.
Now let
\[A = \left( {\begin{array}{*{20}{c}} 1 \\ 2\\ 7 \\ \end{array}} \right)\] and \[B = \left( {\begin{array}{*{20}{c}} 8 \\ 2\\ 1 \\ \end{array}} \right)\]
So now, \[P = A{B^T}\] \[ \Rightarrow P = \left( {\begin{array}{*{20}{c}} 1 \\ 2\\ 7\\ \end{array}} \right)\ \left( {\begin{array}{*{20}{c}} 8 \\ 2\\ 1\\ \end{array}} \right)^{T}\]
To transpose the second matrix, interchange columns and rows. So we get the resultant as,
\[ \Rightarrow P = \left( {\begin{array}{*{20}{c}} 1\\ 2\\ 7\\ \end{array}} \right) \left( {\begin{array}{*{20}{c}} 8&2&1 \\ \end{array}} \right)\]
\[ \Rightarrow P = \left( {\begin{array}{*{20}{c}} 19\\ \end{array}} \right)\] ,which is of order \[1 \times 1\] .
So here we got a non zero matrix, so here \[P\] is a matrix of rank \[1\] , because it has a non zero row in it.

Note:
If we get a null matrix i.e., a matrix in which all elements are zeroes, then it is not said to be a matrix or rank \[1\] . And also make a note that, transpose of a column matrix is a row matrix and similarly, transpose of a row matrix is a column matrix.
Rank of a matrix can be found by converting the matrix into echelon form (upper triangular matrix).
Echelon form \[ = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\ 0&{{a_{22}}}&{{a_{23}}} \\ 0&0&{{a_{33}}} \end{array}} \right)\] , elements below diagonal elements are zeroes.