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Midterm 2-Fall 2006-Solutions

# Midterm 2-Fall 2006-Solutions - AMS 510.01 Fall 2006...

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AMS 510.01 Fall 2006, Midterm #2 — Solutions 1. Indicate whether each of the following statements is true or false. (2 points each). (a) The following matrix, A , is invertible: A = 4 2 1 3 7 2 3 2 0 0 0 1 1 0 0 2 True: We can quickly check the determinant of this matrix, because many of the entries are zero. | A | = 4 2 1 3 7 2 3 2 0 0 0 1 1 0 0 2 = ( - 1) 4 2 1 7 2 3 1 0 0 = ( - 1)(1) 2 1 2 3 = ( - 1)(1)(6 - 2) = 0 (b) 0 2 1 1 1 2 0 3 1 = - 1 False: We take the (21) minor, which requires a negative coefficient. 0 2 1 1 1 2 0 3 1 = ( - 1) 2 1 3 1 = ( - 1)(2 - 3) = +1 (c) F G = G F for any two linear maps F and G . False: Linear maps do not generally commute. Recall that a linear map can be thought of as a matrix, and matrix multiplication is not generally commutative. (d) Every n-square real matrix has n real eigenvalues. False: The eigenvalues may be complex, and may not be unique. (e) A matrix is diagonalizable if and only if it is non-singular. False: There is no relationship between diagonalizability and singularity. (f) < u, v > = < v, u > for all vectors u and v from any real inner product space. True: Conjugate symmetry is a requirement of any inner product, which reduces to simple symmetry for real vectors. (g) The following matrix, B , is diagonalizable: B = 2 - 4 7 - 4 1 5 7 5 3

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True: This is a real symmetric matrix. All such matrices are diagonalizable. (h) An onto mapping can be defined from the set S = { 1 , 2 , 3 } to the set T = { 2 , 4 , 6 , 8 } . False: An onto mapping requires that every element of T have a pre-image. How- ever, there are only three elements in S , so this is not possible. (i) For any non-zero vectors u and v , Proj( u, v ) = 0 if and only if u and v are orthogonal. True: Proj( u, v ) = <u,v> <v,v> v . If v is non-zero, then v <v,v> = 0, thus Proj( u, v ) = 0 iff < u, v > = 0, which is the definition of orthogonality. (j) sgn( { 2 , 3 , 1 } ) = 1. True: The sign of a permutation is determined by the number of swaps required to get the natural sequence. Two swaps are needed ( { 2 , 3 , 1 } → { 1 , 3 , 2 } → { 1 , 2 , 3 } ), so the parity is even, and sgn( { 2 , 3 , 1 } ) = 1. 2. Consider the linear map F : R 3 R 4 defined by: F ( x, y, z ) = ( x + y, x + z, z - x, z - y ); Find the dimension and a basis for the kernel and image of F . (10 points) The kernel of F is the set of vectors from R 3 that map to 0 in R 4 . In other words, it is the solution space of the homogenous system F ( x ) = 0: x + y = 0 x + z = 0 z - x = 0 z - y = 0 We write this in matrix form, and reduce to row echelon form: 1 1 0 0 1 0 1 0 - 1 0 1 0 0 - 1 1 0 1 1 0 0 0 - 1 1 0 0 1 1 0 0 - 1 1 0 1 1 0 0 0 - 1 1 0 0 0 2 0 0 0 0 0 We have no free variables, so the only solution is x = 0. Thus, Ker( F ) = and rank( F ) = dim(Ker( F )) = 0; there is no basis. The image of F is the set of all vectors in R 4 with a pre-image in R 3 . The mappings of any basis of R 3 will span the image. Thus, we start with the usual basis of R 3 , and find a basis for the span of their mappings: F ( e 1 ) = F ((1 , 0 , 0)) = (1 , 1 , - 1 , 0) F ( e 2 ) = F ((0 , 1 , 0)) = (1 , 0 , 0 , - 1) F ( e 3 ) = F ((0 , 0 , 1)) = (0 , 1 , 1 , 1) To find a basis of the span, write the matrix of row vectors and reduce to row echelon form. The rows of any row echelon form are a basis of the span.
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