pset4

# 0 1 1 1 1 0 0 3 1 1 1 1 2 0 1 1 0 2 1 1 1

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 0 C5 0 -C5 -C1 A CA = 0 T 0 0 0 0 0 0 0 0 0 C6 -C6 -C5 K6 = C5 -C6 C6 0 -C2 0 C2 0 0 0 0 + -C 0 C2 0 2 0 0 0 0 0 0 + 0 C4 0 0 -C4 0 0 0 0 -C2 -C3 C2 + C 3 + C 6 -C6 0 0 0 0 0 0 0 0 0 0 0 0 0 0 C6 -C6 -C4 -C5 -C6 0 0 -C4 0 0 + + = If all Ci = 1 0 C4 0 0 0 0 -C5 + 0 C5 -C1 0 -C6 C6 C1 + C2 + C4 -C1 -C2 -C4 C1 + C 3 + C 5 -C3 -C5 C4 + C 5 + C 6 # 3 -1 -1 -1 -1 T 3 -1 -1 A A= -1 -1 3 -1 -1 -1 -1 3 # 7 12) n-1 -1 -1 n-1 K = -1 -1 n-1 -1 -1 n-1 = -1 -1 KK -1 -1 -1 1 n n-1 -1 -1 n-1 K -1 2 1 1 = n 1 1 2 1 1 2 1 1 1 2 1 1 2 2 1 1 2(n - 1) - 1(n - 2) (n - 1) - 2 - (n - 3) -1 + 2(n - 1) - (n - 3) 1 -2 + (n - 1) - (n - 3) = n -2 - (n - 3) + (n - 1) n 1 0 = n 0 1 0 = 0 K= 0 n 0 0 1 0 0 0 0 0 0 0 =I 1 0 0 n -1 - 2 - (n - 4) + (n - 1) # verified n-1 -1 -1 -1 n - 1 -1 -1 -1 n - 1 -1 -1 n-1 -1 -1 -1 n - 1 n - 1 det(n - 1) = n - 1 det n-1 -1 -1 n-1 = (n - 1)2 - 1 = n2 - 2n = n(n - 2) > 0 for n > 2 n-1 det -1 -1 -1 n-1 -1 -1 -1 > 0 n - 1 the upper left determinants > 0 positive definite Alternatively, the eigenvalues of K are = 1, n, n . . . , n > 0 Hence matrix K is positive definite# 8 1 8 17) 7 7 4 1 2 11 2 3 3 9 5 12 8 10 6 4 9 T 6 5 a) Among the 81 entries of A A there are 9 + 12(2) = 33 entries of non-zero values Zero entries in AT A = 81 - 33 = 48# 2 3 4 3 2 0 3 2 0 b) D = c) 2 3 # The middle row has d55 = 4 because node 5 has 4 edges connected to it. There are four -1's in -w because it is next to nodes 2, 4, 6 and 8 # 9...
View Full Document

## This note was uploaded on 05/02/2013 for the course MATH 101 taught by Professor Ns during the Fall '12 term at MIT.

Ask a homework question - tutors are online