Introduction to Linear algebra-Strang-Solutions-Manual_ver13

# 6 0 only two corners 4 0 0 and 0 2 0 let xi 1

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Unformatted text preview: 3 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 off-diagonal 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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