This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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 nsquare 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 nonsingular. 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 preimage. How ever, there are only three elements in S , so this is not possible. (i) For any nonzero 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 nonzero, 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....
View
Full
Document
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.
 Fall '08
 Feinberg,E
 Applied Mathematics

Click to edit the document details