EEL3105_MT2_Solution

EEL3105_MT2_Solution - EEL 3105 Midterm 2 Solution /Fall...

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Unformatted text preview: EEL 3105 Midterm 2 Solution /Fall 2011 14 6 28 6 1. a. AB , (other version AB 26 5 ) 13 5 4 13 6 b. AB 3 3 9 c. AB is not defined as the number of columns of A does not equal the number of rows of B. 2. For A to be singular, det(A). We can compute det(A) by expanding using the 3rd row to give det( A) 1.(4 (4)) (2 (2)) 8 4 0 [The determinant of A for the other version is 4‐2.] Therefore, α=2. 3. a. We know that sum of the eigenvalues is equal to the trace of matrix A, so 3 i 1 i trace( A). Since we are given λ1=13.507, λ2=0.207, the third eigenvalue must be λ3=‐ 0.714. b. We know that the determinant of a matrix is the product of eigenvalues. Therefore, 3 det( A) i 2 i 1 Note that we did not have to calculate the eigenvalues or the determinant explicitly using the data in the matrix A. You could have done the problem without having the entries of A! 4. a. We know that AB and BA will have the same non‐zero eigenvalues. Since BA will be 4×4 matrix, it will have 4 eigenvalues. Therefore, eigenvalues of BA must be 2, 4, ‐2, and 0. [In the other version, the eigenvalues of BA are 1, 2, ‐1, 0.] b. Since determinant of a matrix is the product of eigenvalues, we must have det( BA) 0. c. Trace of a matrix is the sum of eigenvalues. Therefore, trace( BA) 4. (For the other version trace(BA) =2.) d. No BA is singular since its determinant=0. 5. By definition, 1 0 0 AA 0 1 0 . 0 0 1 1 a Now if the vector b is the first column vector of A‐1, we must have 1 1 3 5 a 1 2 4 6 b 0 3 5 8 1 0 We get the following simultaneous equations for a and b: a 3b 5 1 2a 4b 6 0 3a 5b 8 0 We can use any two of these to solve for a and b to get a=‐1, b=‐1. 6. a. Again using the formula that trace equals sum of eigenvalues, we get that the other eigenvalue of A must be 4. b. We know that the determinant is the product of eigenvalues and must therefore be 4. On the other hand, det(A)=6‐b2. Therefore, b2=2 and b= 2. 7. Let’s solve B part first. We are told that A is an nxn matrix whose characteristic polynomial is s n n 1s n 1 n 2 s n 2 1s 0 We start by writing characteristic polynomial of 2A: s s det( sI 2 A) det 2 I A 2n det I A . 2 2 Let us define a new variable z=s/2. [This is a change of variables.] Then det( sI 2 A) 2n det( zI A) 2n [ z n n 1 z n 1 n 2 z n 2 1 z 0 ] n 1 n2 s n s s s 2n n 1 n 2 1 0 2 2 2 2 n n 1 n2 n 1 n 2 s 2 n 1s 2 n 2 s 2 1s 2 0 . The last polynomial is the characteristic polynomial of A. Notice how the coefficients get multiplied by powers of 2. We can apply this general formula for part a. Here n=4 and we have been given the c. p. of A as s 4 3s3 5s 2 7 s 9 . Therefore, c. p. for 2A will be s 4 6s 3 20s 2 56s 144 8. Define M AB BT AT Note that M satisfies M T ( AB BT AT )T BT AT AB M , i. e. M=‐MT. In the solution to the Problem Set 2, I had written that such a matrix is called a skew symmetric matrix as it is the opposite of a symmetric matrix. We are told that is the real eigenvalue of M. Let w be a corresponding eigenvector. Then w 0 and Mw w From here, we need to somehow figure out what l must be. Let’s follow the ideas of the Problem Set 2 solution and also the discussion in class. By multiplying on the left by wT, we get wT Mw wT w wT w. Let us see what we can say about wT Mw. First, we note that wT Mw is a scalar and therefore equals its own transpose. Therefore, wT Mw ( wT Mw) T wT M T w wT Mw. So, it must be the case that wT Mw =0. Therefore, wT w 0 But w 0 and therefor wT w w 0, 2 0 Thus, if is the real eigenvalue of matrix M then it must be 0. In general, a skew symmetric matrix has either purely imaginary or zero eigenvalues. Notice the beautiful structure ‐‐‐ for a symmetric matrix, eigenvalues are always real; for a skew‐symmetric matrix they are always purely imaginary!! ...
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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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