MATH535_HW5

# MATH535_HW5 - Serge Ballif MATH 535 Homework 5 1 a Let A...

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Serge Ballif MATH 535 Homework 5 September 28, 2007 1. a) Let A = ( a ij ) Mat n × m ( ) , such that for each i , n j =1 a ij = 1 . Prove that 1 is an eigen- value of A . b) Show that, if instead n i =1 a ij = 1 for j = 1 , . . . , n , then 1 is still an eigenvalue of A . a) The vector of 1’s v = 1 . . . 1 is an eigenvector for the eigenvalue λ = 1, since A 1 . . . 1 = n j =1 a 1 ,j . . . n j =1 a n,j ! = 1 · v . b) In this case we know that A t satisfies the identity in part a). Hence, A t has 1 as an eigenvalue. We also know that det A = det A t , so χ A = χ A t . Thus, A and A t have the same eigenvalues. Therefore A has eigenvalue 1. 2. Show that there are no proper two-sided ideals I in the ring Mat n × n ( ) when is a field. Hint: Show that if A I Mat n × n ( ) , then elementary row and column operations can be used to show that I n I . Let I Mat n × n ( ) be a nonzero ideal. Let K = ( k ij ) be a nonzero element of I . Denote by E ij the matrix with 1 in the i, j -th slot, and 0 elsewhere. Then the product E ab ( k ij ) E cd is the matrix with k bc in the a, d -th slot and 0 elsewhere. Thus, E ab ( k ij ) E cd = k bc E ad is an element of I . We know that k bc 6 = 0 for some indices b, c . Hence k - 1 bc I n · k bc E ad = E ad is in I . Therefore E 1 , 1 + · · · + E n,n = I n is in I . Therefore

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