C30 For the matrix A below, compute the dimension of the null space of A, dim (N(A)).

$$ \[\begin{bmatrix} 2 & -1 & -3 & 11 & 9 \\ 1 & 2 & 1 & -7 & -3 \\ 3 & 1 & -3 & 6 & 8 \\ 2 & 1 & 2 & -5 & -3 \\ \end{bmatrix}\]

$$

A <- matrix(c(2, -1, -3, 11, 9,
              1, 2, 1, -7, -3,
              3, 1, -3, 6, 8,
              2, 1, 2, -5, -3), 
            nrow = 4, byrow = TRUE)
print(A)
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    2   -1   -3   11    9
## [2,]    1    2    1   -7   -3
## [3,]    3    1   -3    6    8
## [4,]    2    1    2   -5   -3

To find the dimension of the null space of matrix \[A (dim(N(A)))\] we need to perform row reduction (Gaussian elimination) on the augmented matrix \[[A | 0]\]The null space of A is the set of all solutions to the homogeneous system \[Ax = 0\].

# row reduce A
library(pracma)
print(rref(A))
##      [,1] [,2] [,3] [,4] [,5]
## [1,]    1    0    0    1    1
## [2,]    0    1    0   -3   -1
## [3,]    0    0    1   -2   -2
## [4,]    0    0    0    0    0

This matrix indicates that there are three pivot columns (columns 1, 2, and 3) and two free columns (columns 4 and 5) The rank of A is the number of pivot columns, which is 3 So, the dimension of the null space \[(dim(N(A))\] is given by the difference between the total number of columns in A and the rank of A Therefore: Dimension of the null space \[(dim(N(A)) = Total columns − Rank of A\]

Dimension of the null space \[(dim(N(A)) = 5 −3 = 2\] So, the dimension of the null space of matrix = 2