Midterm 2-Fall 2006-Solutions

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

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Unformatted text preview: 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 1 1 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 1 1 2 = (- 1) 4 2 1 7 2 3 1 = (- 1)(1) 2 1 2 3 = (- 1)(1)(6- 2) 6 = 0 (b) 2 1 1 1 2 3 1 =- 1 False: We take the (21) minor, which requires a negative coefficient. 2 1 1 1 2 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 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> 6 = ~ 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 = x + z = z- x = z- y = We write this in matrix form, and reduce to row echelon form: 1 1 1 1- 1 1- 1 1 1 1- 1 1 1 1- 1 1 1 1- 1 1 2 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....
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This note was uploaded on 02/09/2010 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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

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