Vector Spaces - Dimension

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library(knitr)

Problem Statement

The problem C37, selected from page 329 of “A First Course in Linear Algebra”, by Robert A. Bleezer.

Find the rank and nullity of the matrix A =
3 2 1 1 1
2 3 0 1 1
-1 1 2 1 0
1 1 0 1 1
0 1 1 2 -1

Answer: The proof is as follows:
1) A is 5 x 5 square matrix.
2) By Theorem RNNM (page 327), if A is Nonsingular, then rank of A, r(A) = 5 and nullity, n(A) = 0.
3) I claim that A is Nonsingular matrix, which I’ll determine in the below code chunk.

A <- matrix(c(3, 2, -1, 1, 0, 2, 3, 1, 1, 1, 1, 0, 2, 0, 1, 1, 1, 1, 1, 2, 1, 1, 0, 1, -1), nrow = 5)
A

##      [,1] [,2] [,3] [,4] [,5]
## [1,]    3    2    1    1    1
## [2,]    2    3    0    1    1
## [3,]   -1    1    2    1    0
## [4,]    1    1    0    1    1
## [5,]    0    1    1    2   -1

det(A)

## [1] -24

We observe that the Determinant of A = -24 != 0. Therefore A is Nonsingular.

Therefore, the r(A) = 5 and n(A) = 0.

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