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Unformatted text preview: Mathematics of Engineering  ME17 Spring 2007 Homework #5 Solutions Problem 1 [6pts] For matrices, prove that tr( AB ) = tr( BA ) . Given the formula (for nxn matrices A and B with elements A ik and B kj ): tr( AB ) = n X i =1 ( AB ) ii (1) and the fact: ( AB ) ij = n X k =1 A ik B kj (2) Answer: tr( AB ) = n X i =1 ( AB ) ii = n X i =1 n X j =1 ( A ij B ji ) = n X j =1 n X i =1 ( B ji A ij ) = n X j =1 ( BA ) jj = tr( BA ) Explanation: This is true by the definition of the trace of a matrix: tr( AB ) = n X i =1 ( AB ) ii Apply equation (2) to expand ( AB ) ii : n X i =1 ( AB ) ii = n X i =1 n X j =1 A ij B ji Since A ij and B ji are individual entries of the matrices A and B , they commute. (Note: Generally, matrix multiplication does not commute, but scalar multiplication 1 does.) Therefore, we can switch the order of multiplication: n X i =1 n X j =1 A ij B ji = n X i =1 n X j =1 B ji A ij Since we are essentially dealing with a series of individual terms at this step, we can also switch the order of summation: n X i =1 n X j =1 B ji A ij = n X j =1 n X i =1 B ji A ij Now apply the fact given by (2) again (this time in the reverse direction) to get rid of the summation over i : n X j =1 n X i =1 B ji A ij = n X j =1 ( BA ) jj Given the definition of the trace of a matrix (1), notice that the above is equivalent to tr( BA ). Therefore, we have tr( AB ) = tr( BA ) . Problem 2 [10pts] For the matrix A = 3 0 0 2 1 2 0 6 2 Calculate the eigenvalues and eigenvectors for A Answer: Eigenvalues of A: λ 1 = 5 , λ 2 = 2 , λ 3 = 3 Corresponding eigenvectors of A (respectively): v 1 = 1 2 , v 2 =...
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This homework help was uploaded on 04/07/2008 for the course ME 17 taught by Professor Milstein during the Spring '07 term at UCSB.
 Spring '07
 Milstein

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