Problem 1 To get Rank A, of a square matrix make sure that the determinant is not zero, so that we know it is an invertible matrix then Rank = dimension.

(A <- matrix(c(1,-1,0,5,2,0,1,4,3,1,-2,-2,4,3,1,-3), ncol = 4))
##      [,1] [,2] [,3] [,4]
## [1,]    1    2    3    4
## [2,]   -1    0    1    3
## [3,]    0    1   -2    1
## [4,]    5    4   -2   -3
det(A)
## [1] -9

Since determinant of A is not zero then rank = 4

Problem 2 Given an mxn matrix where m > n, what can be the maximum rank? The mini- mum rank, assuming that the matrix is non-zero?

Since m>n, maximum rank is n,

If the matrix had even one non-zero element, its minimum rank would be one.

PROBLEM 3 What is the rank of matrix B?

(A <- matrix(c(1,3,2,2,6,4,1,3,2), ncol = 3))
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
det(A)
## [1] 0

We also notice determinant is zero. so rank has be less than 3 but greater than or equal to one (one or more non zero element) Rank = 1 since row 2 and 3 are multiples of Row 1, We have only one linearly independent row. therefore rank is 1

problem 2 Compute the eigenvalues and eigenvectors of the matrix A. You’ll need to show your work. You’ll need to write out the characteristic polynomial and show your solution.

PLEASE NOTE lambda is denoted at (Ld)

A <- matrix(c(1,0,0,2,4,0,3,5,6), nrow =3)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6

Eigen Values det(A-LdI) =0 A-(Ld)I =

matrix(c('1-(Ld)',0,0,2,'4-(Ld)',0,3,5,'6-(Ld)'), nrow=3)
##      [,1]     [,2]     [,3]    
## [1,] "1-(Ld)" "2"      "3"     
## [2,] "0"      "4-(Ld)" "5"     
## [3,] "0"      "0"      "6-(Ld)"

det(A-(Ld)I) = (1-(Ld))(4-(Ld))(6-(Ld)) 0= 24-34(Ld) + 11(Ld)^2 -(Ld)^3

We can find the roots of this equation using the polyroot function

polyroot(c(24, -34, 11,-1))
## [1] 1+0i 4-0i 6+0i

Eigen Values are below (Ld)1 =1, (Ld)2=4, (Ld)3 =6

Lets find the Eigen vectors (A-(Ld)I)v = 0 for (Ld)1 =1

A-1*(diag(3))
##      [,1] [,2] [,3]
## [1,]    0    2    3
## [2,]    0    3    5
## [3,]    0    0    5

5v3 =0 v3 =0 3v2 + 5(0) =0 v2 =0 We notice that this matrix can not solve for the vector v1 but we can say at v2 =0 and v3=0 (v1,0,0) point, there is an eigen vector or vector for the line 2v2 + 3v3 =0 is the eigen vector

For (Ld)2 =4

A-4*(diag(3))
##      [,1] [,2] [,3]
## [1,]   -3    2    3
## [2,]    0    0    5
## [3,]    0    0    2

2v3 =0 v3 =0 -3v1 +2v2 +3v3 =0 -3v1 +2v2 =0 v2= 3v1/2 = 1.5v1 We notice that this matrix can not solve the 3 vector values we can say for (Ld)2 =4 eigen vector for the plane is -3v1 +2v2 +3v3 =0 and it passes throught the point (0,0,0)

for (Ld)3 =6

A-6*(diag(3))
##      [,1] [,2] [,3]
## [1,]   -5    2    3
## [2,]    0   -2    5
## [3,]    0    0    0

-2v2 + 5v3 =0 v3= (2*v2)/5 = 0.4v2 -5v1 +2v2 + 3v3 =0

-5v1 + 2v2 +3(2*v2/5) =0 v1=16/25(v2) We can say the eigen vector is of the plane -5v1 +2v2 + 3v3 =0, v3=0.4v2 v1=0.64v2 and will have a point in origion (0,0,0)