HW6_Solution - EEL3105 Fall2011 Homework6 Considermatrices...

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EEL 3105 Fall 2011 Homework 6 Consider matrices 12 123 13 4 01 010 001 A k B bb C ccc     1. For A, choose k = 3. Find the eigenvalues of the resulting matrix. Are they real? Why? Solution: The characteristic polynomial of A is 2 det( ) det 5 5 34 IA    It’s roots are given by 5.85 and 0.85 which are the eigenvalues of A. They are real. The reason is that the matrix A is a real symmetric matrix for k=3. Also, note that the sum of eigenvalues is 5 which is the trace of A and their product is 5 which is the determinant of A. 2. Find a formula for eigenvalues of A as a function of k. Suppose k is a design parameter which ranges from 100 to +100. Use the formula to plot the eigenvalues of A as a function of k in the complex plane. Use MATLAB to verify your answer. Such a plot is called root locus. Comment on your results. Solution: Now we will let k remain a design parameter. The characteristic polynomial is 2 det( ) det 5 (4 3 ) 4 I Ak k  It’s roots which are the eigenvalues of A are given by: 59 1 2 2 k  We need to plot these two roots as a function of k. First, let us analyze the behavior of these roots. Their sum will always be 2.5. If 9+12K is non negative, i.e., 0.75 100, k  , then the roots will be real
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adding up to 5. At k=100, the eigenvalues are 14.89 and 19.89. At k= 0.75, the eigenvalues are 2.5 and 2.5, i.e., we have repeated eigenvalues. In problem 1, we saw that for k=3, the eigenvalues are 5.85 and 0.85. At k=0, they are 1 and 4. Thus, as k decreases from 100 to 0.75, the eigenvalues get closer to each other until they meet at 2.5. Now, what happens for the remaining range of k, 100 0.75? k  Well, the first thing we can conclude is that the eigenvalues are complex as the quantity under the square root sign is negative. In this range of k, the eigenvalues are 2.5 ( 9 12 ) / 4 jk  For k= 100, the eigenvalues are 2.5 17.25 j . As k increases from 100 to 0.75, these two complex roots get closer to each other along the vertical direction and at k= 0.75, they meet at 2.5. This completes the picture of the eigenvalues as k varies from 100 to 100. The plot is shown below. Figure 1 We can use MATLAB to make this plot as well. [ Please see my comments on the color code for figures in MATLAB in Prob 4 solution. Here I have played around with the color code! ] k=-100:.01:100; EA=zeros(20000,2); for n=1:1:20000 b1=k(n); A=[1 3;b1 4]; v= eig(A); EA(n,1)=v(1);
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EA(n,2)=v(2); end plot(EA) Figure 2 3. Find characteristic polynomials of B and C in terms of b 1 , b 2 , c 1 , c 2 , and c 2 . Do you see any pattern in your answer? Can you think of a way to generalize this to an nxn matrix?
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This note was uploaded on 12/27/2011 for the course EEL 3105 taught by Professor Boykins during the Fall '10 term at University of Florida.

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HW6_Solution - EEL3105 Fall2011 Homework6 Considermatrices...

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