Ans :- The rank of a matrix is defined as
(i) maximum number of linearly independent vectors columns in the matrix or
(ii) maximum number of linearly independent vectors row in the matrix. Are equal
A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3),nrow=4,ncol=4)
4
## [1] 4
# Calculating rank of Matrix by using rankMatrix function of Matrix package.
require(Matrix)
## Loading required package: Matrix
rankMatrix(A)[1]
## [1] 4
Value is equal, hence Rank of matrix A = 4. It is full matrix and all 4 rows and columns are linearly independent to each other so Rank is 4.
Answer :- a) The maximum Rank for m * n where m>n is n if all n columns are linearly independent.
b) Minimum Rank for m *n where m >n is 1 as matrix is non-zero ; it will have one linear
independent columns.
Answer :- The RANK of matrix B is one as row 2 is equal to row1 multiply by 3 .Row 3 is equal to 2 times row 1. Therefore matrix has only one linear independent row. Rank is 1. In Next step we will calculate the rank of matrix B using rankMatrix function in R to prove it.
B <- matrix (c(1,3,2,2,6,4,1,3,2),nrow=3, ncol=3)
rankMatrix(B)[1]
## [1] 1
As shown above Rank of B is 1.
Answer :- In the attached pdf I will share hand-written solution . Here we prove the same using r functions.
library(pracma)
## Warning: package 'pracma' was built under R version 3.5.3
##
## Attaching package: 'pracma'
## The following objects are masked from 'package:Matrix':
##
## expm, lu, tril, triu
A <- matrix(c(1,0,0,2,4,0,3,5,6),nrow=3,ncol=3)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
charpoly(A)
## [1] 1 -11 34 -24
Above values show that our Charcteristic polynomial is correct.
e <- eigen(A)
e$values
## [1] 6 4 1
Eigen values are correct.
e$vectors
## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0
Eigen vector values are same.