solutions2.3 - Math 307 Problems for section 2.3 1 Let D be...

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Math 307: Problems for section 2.3 October 16, 2012 1. Let D be the incidence matrix in the example done in the course notes. D = 1 1 0 0 0 1 1 0 0 0 1 1 0 1 0 1 1 0 0 1 Using MATLAB/Octave (or otherwise) compute rref ( D ) and find the bases for N ( D ) , R ( D ) and R ( D T ) . Find a basis for N ( D T ) by computing rref ( D T ) . Verify that every loop vector is a linear combination of vectors in this basis.
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>rref(D’) ans = 1 -0 -0 -0 -1 0 1 -0 1 -1 0 0 1 1 -1 0 0 0 0 0 gives the basis N ( D T ) : 0 1 1 1 0 , 1 1 1 0 1 The loop vectors are 1 0 0 1 1 = 0 1 1 1 0 + 1 1 1 0 1 (4) 0 1 1 1 0 = 0 1 1 1 0 (5) 1 1 1 0 1 = 1 1 1 0 1 (6) (7) 2. Draw the graph corresponding to the incidence matrix 1 1 0 0 0 1 1 0 0 0 1 1 1 0 0 1 0 1 0 1 1 0 1 0 . 2
3. How many connected components does the graph whose incidence matrix is in the file hmkgraph.m have (provided on the website)? Explain how you get your answer. 4. Let L = A B be the Laplacian matrix (with L T = L , C invertible and N ( L ) equal to
B C all vectors with constant entries). (a) Show that the voltage to current matrix A B T C 1 B is also symmetric (equal to its transpose). (b) Show that the range of ( A B T C 1 B ) is equal to span { 1 1 } . (Hint: Show that if
(c) Show that (a) and (b) imply that A B T C 1 B is a multiple of 1 1 1 1
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where J = J 1 J 2 . Now J 0 is in R ( L ) and we know that R ( L ) = N ( L ) . So J 0 must be orthogonal to the vectors in the null space of L , and the null space of L is span { 1 1 . . . 1 T } . So J 0 , 1 . . . 1 = J 1 + J 2 = 0 . Therefore the J must be of the form J J for some J R . We also had that J = ( A B T C 1 B ) b for arbitrary b R 2 . Thus all possible vectors J form the range of A B T C 1 B . So the range of A B T C 1 B equals span { 1 1 } . (c) Since R ( A B T C 1 B ) = span { 1 1 } each column must be a scalar multiple of 1 1 . So A B T C 1 B must look like

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