| 1 2 3 4| A =| −1 0 1 3| | 0 1 −2 1| | 5 4 −2 −3|
Convert A to RREF
| 1 2 3 4| RREF | 1 0 0 0| A =| −1 0 1 3| ——> | 0 1 0 0| | 0 1 −2 1| | 0 0 1 0| | 5 4 −2 −3| | 0 0 0 1|
rank is 4, there are no linear independent rows
A = matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), nrow=4, byrow = TRUE)
A
## [,1] [,2] [,3] [,4]
## [1,] 1 2 3 4
## [2,] -1 0 1 3
## [3,] 0 1 -2 1
## [4,] 5 4 -2 -3
pracma::rref(A)
## [,1] [,2] [,3] [,4]
## [1,] 1 0 0 0
## [2,] 0 1 0 0
## [3,] 0 0 1 0
## [4,] 0 0 0 1
Matrix:: rankMatrix(A)
## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16
maximum rank would be m as it is possible as in problem set one that no rows are linearly independent. Tre minimum rank is 1
What is the rank of matrix B?
|1 2 1| RREF |1 2 1|
B = |3 6 3| ——> |0 0 0| |2 4 2| |0 0 0|
B = matrix(c(1,2,1,3,6,3,2,4,2),nrow = 3, byrow = TRUE)
B
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 3 6 3
## [3,] 2 4 2
pracma::rref(B)
## [,1] [,2] [,3]
## [1,] 1 2 1
## [2,] 0 0 0
## [3,] 0 0 0
Matrix::rankMatrix(B)
## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16
####2 Problem set 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
|1 2 3|
A = |0 4 5|
|0 0 6|
| λ 0 0|
λI3 =| 0 λ 0|
| 0 0 λ|
| λ-1 -2-3|
λI3 = -A =| 0 λ -4-5|
| 0 0 λ-6|
apply rule of Sarrus
| λ-1 -2 -3 | λ-1 -2
λI3 = -A =| 0 λ -4 -5 | 0 λ-4
| 0 0 λ -6 | 0 0
(λ-1)(λ-4)(λ-6)+(-2)(-5)(0)+(-3)(0)(0)-(-2)(0)(λ-6)- -(λ-1)(-5)(0)-(-3)(λ-4)(0)
everything drops out except the first terms
(λ-1)(λ-4)(λ-6)
this leaves us with three eingenvalues
λ = 1
λ = 4
λ = 6
eigenvector for
λ = 1
plug value in for lambda
|0 -2 -3| RREF |0 1 0||v1| |0|
λI3 = -A = |0 -3 -5| ——> |0 0 1||v2| = |0| |0 0 -5| |0 0 0||V3| |0| v1 = t v2 = 0 v3 = 0
|1|
E1 = t |0|
|0|
eigenvector for
λ=4
plug value in for lambda
|3 -2 -3| RREF |1 2/3 0||v1| |0|
λI3 = -A = |0 0 -5| ——> |0 0 1||v2| = |0| |0 0 2| |0 0 0||v3| |0|
v1 + 2/3v2 = 0
v3 = 0
eigenvector for
λ = 6
plug value in for lambda
|5 -2 -3| RREF |1 0 -1.6||v1| |0|
λI3 = -A = |0 2 -5| ——-> |0 1 -2.5||v2|=|0| |0 0 0| |0 0 0 ||v3| |0|
v1 - 1.6v3 = 0 -> 0.625v1 = v3
v2 - 2.5v3 = 0 -> 0.4v2 = v3
v3 = t
|0.625|
E6 = t|0.4 |
|0 |
A = matrix(c(1,2,3,0,4,5,0,0,6), nrow = 3, byrow = TRUE)
A
## [,1] [,2] [,3]
## [1,] 1 2 3
## [2,] 0 4 5
## [3,] 0 0 6
eigen(A)
## eigen() decomposition
## $values
## [1] 6 4 1
##
## $vectors
## [,1] [,2] [,3]
## [1,] 0.5108407 0.5547002 1
## [2,] 0.7981886 0.8320503 0
## [3,] 0.3192754 0.0000000 0
B = matrix(c(0,-2,-3,0,-3,-5,0,0,-5), nrow = 3, byrow = TRUE)
B
## [,1] [,2] [,3]
## [1,] 0 -2 -3
## [2,] 0 -3 -5
## [3,] 0 0 -5
pracma::rref(B)
## [,1] [,2] [,3]
## [1,] 0 1 0
## [2,] 0 0 1
## [3,] 0 0 0
C = matrix(c(3,-2,-3,0,0,-5,0,0,0), nrow =3, byrow = TRUE)
C
## [,1] [,2] [,3]
## [1,] 3 -2 -3
## [2,] 0 0 -5
## [3,] 0 0 0
pracma::rref(C)
## [,1] [,2] [,3]
## [1,] 1 -0.6666667 0
## [2,] 0 0.0000000 1
## [3,] 0 0.0000000 0
D = matrix(c(5,-2,-3,0,2,-5,0,0,0), nrow = 3, byrow = TRUE)
D
## [,1] [,2] [,3]
## [1,] 5 -2 -3
## [2,] 0 2 -5
## [3,] 0 0 0
pracma::rref(D)
## [,1] [,2] [,3]
## [1,] 1 0 -1.6
## [2,] 0 1 -2.5
## [3,] 0 0 0.0