Problem_Set2_Solution - EEL 3105 Fall 2011 Problem Set 2 1....

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Unformatted text preview: EEL 3105 Fall 2011 Problem Set 2 1. Consider the vector v [1 2.8 3 3.33] Find vvT. Is it symmetric? Find its eigenvalues. Is it invertible? Solution: We can calculate vvT directly. It is a scalar and its value is 28.92. Since it is a scalar, it is symmetric. Again, since it is a scalar, its (only eigenvalue) is 28.92. Since this eigenvalue is nonzero, it is invertible. Although, I did not ask for this, let us consider vTv. This is the 4x4 matrix: 1.0000 2.8000 ‐3.0000 3.3300 2.8000 7.8400 ‐8.4000 9.3240 ‐3.0000 ‐8.4000 9.0000 ‐9.9900 3.3300 9.3240 ‐9.9900 11.0889 What if I had asked find eigenvalue of vTv? What would you do? Would you go about writing characteristic polynomial of this 4x4 matrix? If so, that would be a long and tedious process. And it would certainly not work for Problem2. On the other hand, we can go back to the last homework where we showed any nonzero eigenvalue of AB is also a nonzero eigenvalue of BA. Now, let us set A to be vT and B to be v. Then BA has only one eigenvalue, i. e., 28.92. Therefore, AB which is the above 4x4 matrix must have 28.92 as its eigenvalue. Can AB= vTv have any other non‐zero eigenvalue? Well, if it did then that non‐zero eigenvalue of vTv T T would be an eigenvalue of vv but vv has only one eigenvalue, 28.92. Thus, all other eigenvalues of vTv must be zero. Thus, the only eigenvalue of vTv are 28.92 and zero. In fact, vTv has a triple eigenvalue at zero. 2. Let v be an nx1 vector. Find eigenvalues of vvT. For choices of v is vvT invertible? Solution: First, if v is the zero vector, then vvT is the zero matrix and its eigenvalues are all zero and it is not invertible. Now, suppose v is a non‐zero nx1 vector, vvT is a nxn matrix. To find its eigenvalues, we first see that vTv is a non‐zero scalar and is the one and only eigenvalue of vTv. Now, we know that nonzero eigenvalues of a matrix product AB and BA are the same. With A=v and B=vT, we can use this fact to conclude that the only non‐zero eigenvalue of vvT is vTv and the rest must be zero. For any matrix to be invertible, all its eigenvalues must be non‐zero. Therefore, the only way vvT can be invertible is if n=1 and v is a non‐zero scalar. Another way of seeing this is as follows. Suppose M is inverse of vvT. Then MvvT = vvTM= I. Let vT M w [ w1 w2 wn ] Now vvT M vw v w1 w2 wn vw1 vw2 vwn T Therefore, each column of vv M is a scalar multiple of v. But these columns are also columns of the identity matrix. But j‐th column of I is all zero except for 1 in the j‐th spot. Thus, any two columns of I cannot be scalar multiples of a fixed column unless n=1. Hence, n cannot bigger than 1. 3. Let A and B be two square nxn matrices. Find a formula for (A+B)2. Solution: ( A B) 2 ( A B)( A B) AA AB BA BB A2 AB BA B 2 Please note that we cannot equate the last term to A2 2 AB B 2 ! The only time this would happen is if AB=BA, i. e. A and B commute. 4. Suppose AB=BA. Find a formula for (A+B)n. What will happen if A and B do not commute? Solution: Since AB = BA, we can hope that the binomial formula will give us the right answer. First let us recall the binomial formula: n n ( x y ) n x j y n j j 0 j So, a guess would be: n n ( A B)n A j B n j j 0 j The easiest proof is by induction. This is clearly true for n=2. Suppose it is true for n. We then need to prove it for n+1. Here is how we might do it. n n ( A B)( n 1) ( A B )( A B) n ( A B) A j B ( n j ) j 0 j n n n n A j 1 B ( n j ) A j B ( n j 1) j 0 j j 0 j n 1 n n n A( n 1) A( j 1) B ( n j ) A j B ( n j 1) B ( n 1) j 0 j j 1 j n n n k ( n 1 k ) ( n 1) ( n 1) B A A B k 1 k 1 k n n 1 k ( n 1 k ) A( n 1) B ( n 1) A B k k 1 n 1 n 1 k ( n 1 k ) A B k 0 k This shows that the formula holds for (n+1). Thus the proof is complete using induction. You should go through each step and see if you can follow how I went from one step to the next. Key is to break the summations as indicated. I also used the so‐called Pascal’s identity n n n 1 k k 1 k which is a key property in establishing the binomial theorem. If A and B do not commute, then (A+B)n will be sum of all possible products of A and B so that A occurs exactly k times and B occurs (n‐k) times for k = 0, 1, 2, …,n. For n=2, we saw the answer above. You can try n=3. For general n, we will get a similar formula but there will be lot more combinations. 5. Suppose the characteristic polynomial of a 3x3 matrix A is s 3 4 s 2 5s 3 Find the characteristic polynomial of I+A where I is a 3x3 identity matrix. Solution: See the solution to the next problem where I am going to solve the general problem. After you read that solution, come back and read the rest. Using that solution, we will get: det( sI ( I A)) ( s 1)3 4( s 1) 2 5( s 1) 3 s 3 3s 2 3s 1 4 s 2 8 s 4 5 s 5 3 s3 s 2 1 Let’s calculate roots of the two c. p. to find eigenvalues of A and I+A. Using MATLAB: >> f=[1 4 5 3]; >> roots(f) ans = ‐2.4656 ‐0.7672 + 0.7926i ‐0.7672 ‐ 0.7926i >> g=[1 1 0 1]; >> roots(g) ans = ‐1.4656 0.2328 + 0.7926i 0.2328 ‐ 0.7926i You can see that the eigenvalues of (I+A) are exactly the eigenvalues of A with 1 added to each! 6. Suppose A is an nxn matrix with characteristic polynomial s n n 1s n 1 n 2 s n 2 1s 0 Find the characteristic polynomial of I+A where I is an nxn identity matrix. Solution: Notation: here I have used s as the variable instead of . So, I will stick to s. Let us start with the definition of characteristic polynomial and use it to find the c. p. for I+A: det( sI ( I A)) det( sI I A) det(( s 1) I A) Now note that the last expression is highly suggestive. Define z=s‐1. Then, we can get det( sI ( I A)) det( zI A) z n n 1 z n 1 ... 1 z 0 ( s 1) n 1 ( s 1) ... 1 ( s 1) 0 n n The last polynomial is the c. p. of I+A. Now, suppose is an eigenvalue of A. Then, by definition, n n 1 n 1 ... 1 0 0 It follows that det(( 1) I ( I A)) ( 1 1) n n 1 ( 1 1) n 1 ... 1 ( 1 1) 0 n n 1 n 1 ... 1 0 0 Therefore, (+1) is an eigenvalue of I+A. Conversely, if w is an eigenvalue of I+A, then w‐1 is an eigenvalue of A. Thus, if we knew eigenvalues of A, we could compute eigenvalues of I+A with extreme ease. Now go back and read the solution to Problem 5. 7. Let A and B be two matrices such that A is nxm and B is mxn. Is AB+BTAT a symmetric matrix? Please justify your answer. Solution: Let M= AB+BTAT. Let’s see what is MT: M T ( AB BT AT )T ( AB)T ( BT AT )T BT AT AB M TT Thus, M is a symmetric matrix. Therefore, AB+B A is symmetric. TT Just for fun: What about AB‐ B A ? Find out the answer for yourself. When a matrix M is such that M=‐ MT, it is called skew‐symmetric. Take an example of a skew symmetric matrix and use MATLAB to compute its eigenvalues. 8. Let A be a symmetric matrix. Suppose 1+jb is an eigenvalue of A. Find b. Solution: Since A is a symmetric (real) matrix, its eigenvalues are guaranteed to be real. Therefore, b must be zero. 9. Consider the matrix 1 b A b 2 Let v = [1 ] be a 1x2 row vector. A. Calculate vAvT. B. Set =‐1. Under what conditions on b is vAvT positive? C. Find eigenvalues of A. Under what conditions on b are both eigenvalues positive? Solution: A: We use definitions and calculate: 1 b 1 1 b 1 b 2 b 2 2 1 b b 2 1 1 2b 2 2 B. Suppose =‐1. Then vAv =3‐2b. Therefore, so long as b 1.5, vAv will be positive. T T C. Let’s calculate the characteristic polynomial of A: 1 b det( I A) det b 2 ( 1)( 2) b 2 2 3 2 b 2 The roots of c. p. are 3 9 4(2 b 2 ) 3 1 4b 2 2 2 First thing we see is that the eigenvalues are real. This had to be true since A is real symmetric. Now, for the eigenvalues to be positive, 1 4b 2 3 1 4b 2 9 b 2 2 2 b 2 . This is the range of b for which the eigenvalues of A are positive. 10. Let 1 2 A 4 5 3 1 Find eigenvalues of B= ATA. Are they positive? Find the corresponding eigenvectors w1 and w2. Calculate T T T T w1 Bw1 and w2 Bw2 . Calculate their ratio to w1 w1 and w2 w2 respectively. Comment on your answer. Solution: We start by computing B: 1 2 1 4 3 26 21 B A A 4 5 21 30 2 5 1 3 1 T Note that B is a real symmetric matrix. This should have been expected from definition of B as ATA. Characteristic polynomial of B is: s 26 21 det( sI B) det 21 s 30 ( s 26)( s 30) 441 s 2 56s 339 Eigenvalues of B are roots of this quadratic polynomial which are: 49.095 and 6.905. To find eigenvectors, we set up the eigenvector equations: p p p w1 ; B 49.095 q q q 26 p 21q 49.095 p 21 p 30q 49.095q e e e w2 ; B 6.905 f f f 26e 21 f 6.905e 21e 30 f 6.905 f Solutions for p,q and e,f are non‐unique since the equations are not independent. As you will recall from class lectures, this will always happen for eigenvector equations. We can get a pair of solutions: p 1 q 1.0998 e 1.0998 f 1 Note that w1 and w2 are orthogonal or perpendicular as they are guaranteed to be because of symmetry of B! We could have made these into unit vectors by dividing by their magnitude or norm. I have not done so as it was not required. T T Finally, let us calculate w1 Bw1 and w2 Bw2 . This is a straightforward numerical calculation. The results are: Now, the ratios: 26 21 1 T w1 Bw1 1 1.0998 108.478 21 30 1.0998 26 21 1.0998 T w2 Bw2 1.0998 1 15.257 21 30 1 T w1 Bw1 108.478 49.095 T w1 w1 2.209 T w2 Bw2 15.257 6.905 T w2 w2 2.209 Note that the ratios are precisely the eigenvalues!! Should this be a surprise? Not at all! Note that Bw1 49.095w1 T T w1 Bw1 49.095w1 w1 T w1 Bw1 49.095 T w1 w1 11. Suppose A is an nxm matrix. Set B = ATA. Suppose is an eigenvalue of A. Suppose w is an eigenvector of B corresponding to the eigenvalue . Find a formula for wTBw. Can you use this to show that cannot be negative? Solution: This is a general version of Problem 10. Bw w wT Bw wT w Now note that wT Bw wT AT Aw || Aw ||2 wT w || w ||2 wT Bw || Aw ||2 0 wT w || w ||2 Thus, we see that eigenvalues of a matrix of the form ATA are always real and positive!! We saw this numerically in Problem 11. But it is a very general property. Recall such matrices occur in least squares calculations among many other places. ...
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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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