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1
2
1
5=4
c
1
2
1
3
2;fD
0 yields x D
1 C any c ;
7 AT
AD
2
2
2
4
1
7=8
c
3
potentials x D 5 ; 1; 7 and currents CAx D 1 ; 4 ; 1 .
4
8
4
4
2
3
23
23
23
1
1
00
1
0
1
61
6 17
6 07
0
1 07
6
7
617
67
67
1
1 0 7 leads to x D 4 5 and y D 6 1 7 and 6 1 7 solving
8 AD6 0
1
40
5
4 05
4 15
1
01
1
0
0
11
0
1
AT y D 0 .
9 Elimination on Ax D b always leads to y T b D 0 in the zero rows of U and R: b1 C b2 b3 D 0 and b3 b4 C b5 D 0 (those y ’s are from Problem 8 in the left
nullspace). This is Kirchhoff’s Voltage Law around the two loops.
2
3
1100
The nonzero rows of U keep
6 0 1 1 07
6
7 edges 1, 2, 4. Other spanning trees
10 The echelon form of A is U D 6 0 0 1 1 7
4 0 0 0 0 5 from edges, 1, 2, 5; 1, 3, 4; 1, 3, 5;
1, 4, 5; 2, 3, 4; 2, 3, 5; 2, 4, 5.
0000
2
3
diagonal entry D number of edges into the node
2
1
1
0
3
1
1 7 the trace is 2 times the number of nodes
61
T
11 A A D 4
1
1
3
1 5 offdiagonal entry D 1 if nodes are connected
0
1
1
2
AT A is the graph Laplacian, AT CA is weighted by C 12 (a) The nullspace and rank of AT A and A are always the same (b) AT A is always
positive semideﬁnite because x A Ax D kAx k 0. Not positive deﬁnite because
rank is only 3 and .1; 1; 1; 1/ is in the nullspace (c) Real eigenvalues all 0 because
positive semideﬁnite.
T T 2 Solutions to Exercises 84
2 4
62
T
13 A CAx D 4
2
0 2
8
3
3 2
3
8
3 3
2
0
37
6
xD4
35
6 3
51
1
gives four potentials x D . 12 ; 6 ; 1 ; 0/
6
07
I grounded x4 D 0 and solved for x
05
2
currents y D CAx D . 2 ; 3 ; 0; 1 ; 1 /
1
3
22 14 AT CAx D 0 for x D c.1; 1; 1; 1/ D .c; c; c; c/. If AT CAx D f is solvable, then f in
15
16 17
18 the column space (D row space by symmetry) must be orthogonal to x in the nullspace:
f 1 C f 2 C f 3 C f 4 D 0.
The number of loops in this connected graph is n m C 1 D 7 7 C 1 D 1.
What answer if the graph has two separate components (no edges between)?
Start from (4 nodes) (6 edges) C (3 loops) D 1. If a new node connects to 1 old
node, 5 7 C 3 D 1. If the new node connects to 2 old nodes, a new loop is formed:
5 8 C 4 D 1.
(a) 8 independent columns (b) f must be orthogonal to the nullspace so f ’s add
to zero (c) Each edge goes into 2 nodes, 12 edges make diagonal entries sum to 24.
A complete graph has 5 C 4 C 3 C 2 C 1 D 15 edges. With n nodes that count is
1 C C .n 1/ D n.n 1/=2. Tree has 5 edges. Problem Set 8.3, page 437
1 Eigenvalues D 1 and .75; (A Ax D x .
:6
2 AD
:4 1
1 1
:75 1
:4 3 D 1 and :8, x D .1; 0/; 1 and I /x D 0 gives the steady state x D .:6; :4/ with
1
:6
; A1 D
:6
:4 1
1 1
0 0
0 1
:4
1
:6 :6
D
.
:6
:4 :4 5
1
1
:8, x D . 9 ; 4 /; 1; 1 , and 1 , x D . 3 ; 1 ; 3 /.
9
4
4
3 4 AT always has the eigenvector .1; 1; : : : ; 1/ for D 1, because each row of AT adds to 1. (Note again that many authors use row vectors multiplying Markov matrices.
So they transpose our form of A.)
5 The steady state eigenvector for D 1 is .0; 0; 1/ D everyone is dead.
6 Add the components of Ax D x to ﬁnd sum s D s . If ¤ 1 the s...
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 Spring '12
 Minki
 Mass

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