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Unformatted text preview: Solutions to the selected problems (Homework 8,9) Linear Algebra Fall 2010 Page 163, 8) Since T B ( I ) = IB BI = 0, the nullspace of T B is not the zero space, therefore T B is singular and not invertible, so det T B = 0. Page 190, 8) To show that I BA is invertible, it is enough to show that ( I BA )( I + B ( I AB ) 1 A ) = I. (then we can conclude that ( I + B ( I AB ) 1 A )( I BA ) = I , and so I BA is invertible). This is shown as follows: ( I BA )( I + B ( I AB ) 1 A ) = I BA + B ( I AB ) 1 A BAB ( I AB ) 1 A = I BA + B ( I AB )( I AB ) 1 ) A = I BA + BA = I. Page 190, 6) The matrix is lower triangular with all the diagonal entries zero, so the only characteristic value for the matrix is zero. If it is diagonalizable, then the nullspace: { X  AX = 0 } should be four dimensional, so every vector in R 4 × 1 should be a in the above set which means A = 0. So A is diagonalizable if and only if a = b = c = 0. Page 190, 7) This is an immediate corollary of the lemma on page 186....
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.
 Spring '08
 Neely

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