MA122 Lab Report 10
Name: Student Number: Spring 2021
1. [3 marks] Recall
2 Question #1, Lab 4,
3 where the Google PageRank algorithm was discussed and each entry aij in the standard
1=4 0 1 1=2
6 1=4 0 0 0 7
matrix A = 6 7
4 1=4 1=2 0 1=2 5 represented how much webpage j "endorsed" webpage i in an internet of 4 webpages.
1=4 1=2 0 0
(a) Given that ! is an eigenvector of A; evaluate A!
T T
v = v1 v2 v3 v4 = 8=3 2=3 3=2 1 v and use the
result to …nd the corresponding eigenvalue : [Note: Do not convert to decimals. Leave results as exact values.]
2 32 3 2 3
1=4 0 1 1=2 8=3 8=3
6 1=4 0 0 0 7 6 7 6 2=3 7
A! v =6 7 6 2=3 7=6 7 ! !
4 1=4 1=2 0 1=2 5 4 3=2 5 4 3=2 5 ) A v = v when = 1 (i.e. = 1 is corresponding eigenvalue)
1=4 1=2 0 0 1 1
!
(b) The "billion dollar eigenvector $ " is what Google uses for its PageRank algorithm to rank webpages in a search (as
opposed to the matrix multiplication we did in Lab Report 4). In this example, the billion dollar eigenvector would be
1
! P4
! !
$ = vi v (i.e. a scalar multiple of !
v whose entries sum to 1). Find $ .
i=1
3 2 2 3 2 3
8=3 8=3 16=35
! 1 6 2=3 7 6 2=3 7 6 7
$ = 6 7= 6 6 7 = 6 4=35 7
8=3 + 2=3 + 3=2 + 1 4 3=2 5 35 4 3=2 5 4 9=35 5
1 1 6=35
!
[Note: Row i with the largest entry in $ would be the webpage ranked …rst in a Google search and so on. Who knew
an eigenvector could have made you billions of dollars?! I guess they are useful.]
1 3
2. [9 marks] Consider the matrix A = :
3 9
(a) Determine (by hand) 1 and 2; the two eigenvalues of A:
+1 3 2
det ( I2 A) = 0 ) = ( + 1) ( + 9) 9= + 10 = ( + 10) = 0
3 +9
) 1 = 0 and 2 = 10 are the eigenvalues of A:
(b) For each eigenvalue in part (a), …nd (by hand) the corresponding eigenvectors of A.
1 3 0 1 3 0
For 1 = 0 : (0I2 A) ~v1 = ~0 )
3 9 0 0 0 0
3
) the eigenvectors corresponding to 1 are t1 (or Span([3 1]T )) where t1 is non-zero.
1
9 3 0 1 1=3 0 1 1=3 0
For 1 = 10 : ( 10I2 A) ~v1 = ~0 )
3 1 0 3 1 0 0 0 0
1=3
) the eigenvectors corresponding to 2 are t2 (or Span([ 1=3 1]T )) where t2 is non-zero.
1
1
(c) State matrix P that diagonalizes A and determine P . Then (by hand) use the result to …nd A6 :
3 1=3 1 1 1 1=3 3=10 1=10
P = and P = =
1 1 3(1) ( 1=3)(1) 1 3 3=10 9=19
3 1=3 0 0 3=10 1=10 0 1000000=3 3=10 1=10 100000 300000
A6 = P D 6 P 1
= = =
1 1 0 ( 10)6 3=10 9=10 0 1000000 3=10 9=10 300000 900000
This study source was downloaded by 100000858061865 from CourseHero.com on 01-15-2023 14:02:26 GMT -06:00
https://www.coursehero.com/file/102421453/MA122Lab10Solutionspdf/
Name: Student Number: Spring 2021
1. [3 marks] Recall
2 Question #1, Lab 4,
3 where the Google PageRank algorithm was discussed and each entry aij in the standard
1=4 0 1 1=2
6 1=4 0 0 0 7
matrix A = 6 7
4 1=4 1=2 0 1=2 5 represented how much webpage j "endorsed" webpage i in an internet of 4 webpages.
1=4 1=2 0 0
(a) Given that ! is an eigenvector of A; evaluate A!
T T
v = v1 v2 v3 v4 = 8=3 2=3 3=2 1 v and use the
result to …nd the corresponding eigenvalue : [Note: Do not convert to decimals. Leave results as exact values.]
2 32 3 2 3
1=4 0 1 1=2 8=3 8=3
6 1=4 0 0 0 7 6 7 6 2=3 7
A! v =6 7 6 2=3 7=6 7 ! !
4 1=4 1=2 0 1=2 5 4 3=2 5 4 3=2 5 ) A v = v when = 1 (i.e. = 1 is corresponding eigenvalue)
1=4 1=2 0 0 1 1
!
(b) The "billion dollar eigenvector $ " is what Google uses for its PageRank algorithm to rank webpages in a search (as
opposed to the matrix multiplication we did in Lab Report 4). In this example, the billion dollar eigenvector would be
1
! P4
! !
$ = vi v (i.e. a scalar multiple of !
v whose entries sum to 1). Find $ .
i=1
3 2 2 3 2 3
8=3 8=3 16=35
! 1 6 2=3 7 6 2=3 7 6 7
$ = 6 7= 6 6 7 = 6 4=35 7
8=3 + 2=3 + 3=2 + 1 4 3=2 5 35 4 3=2 5 4 9=35 5
1 1 6=35
!
[Note: Row i with the largest entry in $ would be the webpage ranked …rst in a Google search and so on. Who knew
an eigenvector could have made you billions of dollars?! I guess they are useful.]
1 3
2. [9 marks] Consider the matrix A = :
3 9
(a) Determine (by hand) 1 and 2; the two eigenvalues of A:
+1 3 2
det ( I2 A) = 0 ) = ( + 1) ( + 9) 9= + 10 = ( + 10) = 0
3 +9
) 1 = 0 and 2 = 10 are the eigenvalues of A:
(b) For each eigenvalue in part (a), …nd (by hand) the corresponding eigenvectors of A.
1 3 0 1 3 0
For 1 = 0 : (0I2 A) ~v1 = ~0 )
3 9 0 0 0 0
3
) the eigenvectors corresponding to 1 are t1 (or Span([3 1]T )) where t1 is non-zero.
1
9 3 0 1 1=3 0 1 1=3 0
For 1 = 10 : ( 10I2 A) ~v1 = ~0 )
3 1 0 3 1 0 0 0 0
1=3
) the eigenvectors corresponding to 2 are t2 (or Span([ 1=3 1]T )) where t2 is non-zero.
1
1
(c) State matrix P that diagonalizes A and determine P . Then (by hand) use the result to …nd A6 :
3 1=3 1 1 1 1=3 3=10 1=10
P = and P = =
1 1 3(1) ( 1=3)(1) 1 3 3=10 9=19
3 1=3 0 0 3=10 1=10 0 1000000=3 3=10 1=10 100000 300000
A6 = P D 6 P 1
= = =
1 1 0 ( 10)6 3=10 9=10 0 1000000 3=10 9=10 300000 900000
This study source was downloaded by 100000858061865 from CourseHero.com on 01-15-2023 14:02:26 GMT -06:00
https://www.coursehero.com/file/102421453/MA122Lab10Solutionspdf/
