Assignment 3

Problem Set 1

1.1)

A <- matrix(c(1,2,3,4,-1,0,1,3,0,1,-2,1,5,4,-2,-3), byrow=T, nrow=4, ncol=4)
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
library(Matrix)
## Warning: package 'Matrix' was built under R version 3.3.3
rankMatrix(A)
## [1] 4
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 8.881784e-16
#another way to compute rank
qr(A)
## $qr
##            [,1]       [,2]      [,3]      [,4]
## [1,] -5.1961524 -4.2339020  1.539601  2.694301
## [2,] -0.1924501 -1.7533038 -1.436442 -4.795180
## [3,]  0.0000000  0.5703518  3.683241  2.162187
## [4,]  0.9622504 -0.5877259  0.592032  0.268209
## 
## $rank
## [1] 4
## 
## $qraux
## [1] 1.192450 1.573827 1.805914 0.268209
## 
## $pivot
## [1] 1 2 3 4
## 
## attr(,"class")
## [1] "qr"

The rank of matrix A is 3.

1.2)

\[rank(A_{mxn}) <= min(m,n)\] (m,n) denotes the smaller of the two numbers m and n, or either m or n if m=n. The rank of the matrix can be no larger than this smaller value. For example, for a 3 by 5 matrix, the rank can be no larger than 3.

Therefore, given an mxn matrix where m > n, the maximum rank is n. Assuming the matrix is non-zero, the minimum rank is 1.

1.3)

B <- matrix(c(1,2,1,3,6,3,2,4,2), byrow=T, nrow=3, ncol=3)
B
##      [,1] [,2] [,3]
## [1,]    1    2    1
## [2,]    3    6    3
## [3,]    2    4    2
rankMatrix(B)
## [1] 1
## attr(,"method")
## [1] "tolNorm2"
## attr(,"useGrad")
## [1] FALSE
## attr(,"tol")
## [1] 6.661338e-16
#another way to compute rank
qr(B)
## $qr
##            [,1]      [,2]      [,3]
## [1,] -3.7416574 -7.483315 -3.741657
## [2,]  0.8017837  0.000000  0.000000
## [3,]  0.5345225  0.000000  0.000000
## 
## $rank
## [1] 1
## 
## $qraux
## [1] 1.267261 0.000000 0.000000
## 
## $pivot
## [1] 1 2 3
## 
## attr(,"class")
## [1] "qr"

The rank of matrix B is 1.

Problem Set 2

2.1

A <- matrix(c(1,2,3,0,4,5,0,0,6), byrow=T, nrow=3, ncol=3)
A
##      [,1] [,2] [,3]
## [1,]    1    2    3
## [2,]    0    4    5
## [3,]    0    0    6
A_eigen <- eigen(A)
A_eigen
## $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

\[A=\begin{vmatrix}1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6\end{vmatrix}\] We need to find eigenvalues from the characteristic polynomial. First, we’ll subtract \[\lambda\] from the diagonal entries in the matrix.

\[\begin{vmatrix}1-\lambda & 2 & 3 \\ 0 & 4-\lambda & 5 \\ 0 & 0 & 6-\lambda\end{vmatrix}\] \[-\lambda+11\lambda^2-34 \lambda+24=0\] \[(-1)(\lambda-1)(\lambda-4)(\lambda-6)=0\] The eigenvalues and the associated eigenvectors for matrix A are:

\[\lambda_1 = 1\] \[\begin{vmatrix}1\\0\\0\end{vmatrix}\] \[\lambda_2 = 4\] \[\begin{vmatrix}1\\ \frac{3}{2} \\0\end{vmatrix}\]

\[\lambda_3 = 6\] \[\begin{vmatrix}1\\ \frac{25}{16} \\\frac{5}{8}\end{vmatrix}\]