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Unformatted text preview: Answer Key to Second Midterm Examination, Version 1 1. Let V be a two dimensional subspace of 5 . Answer the following questions (a correct answer without step by step description of your reasoning: maximum 10 points deduction each): (a) Find the dimension of V . x1 x2 x5 x can be considered to be in V and the right x can be considered to be in V . Since x2 0 for all j 1 2 5, the only possibility for x j to have x2 x2 x2 0 is j 1 2 5 x1 x2 x5 0. Thus x 0 and V V 0 . (c) For two distinct orthonormal basis v 1 v 2 and u 1 u 2 of V and any given v1 x v1 v 2 x v 2 and w u1 x u1 u2 vector x in 5 , let v x u 2 . Prove that v w . Hint: Show first that x v is in V , and use this fact and (b) to show v w. because v i v j 0 if i j and v j v j 1. This shows x v is in V . In the same way, we find that x w is in V , replacing in the above formula, v j by u j and v by w . Take the difference v w which is in V . Since v w x w x v , w are in V . Thus v w is in V V 0 by (b), which shows we find that v that v w. 2. Compute the following numbers and vectors (a correct answer without step by step description of computation: 10 points deduction each). a. The rank of A. By row operation, we get b. The nullity of A. The sum of the nullity and the rank is the dimension of the source; so, The nullity is 1.
1 Thus rankA 2. " 3 4 1 3 1 3 4 4 4 4 4 4 1 3 1 3 1 2 Let A 1 7 10 4 2 8 11 5 3 9 12 6 . 1 7 10 4 2 8 11 5 3 9 12 6 1 2 3 0 6 12 0 9 18 0 3 6 1 0 0 0 2 1 1 1 3 2 2 2 1 0 0 0 0 1 1 2 0 0 0 0 j j ! x v x v j v 1 v v 1 v j v 2 v v 2 ) & 0 % ' ( % ' x v v j ! We compute for j 1 2, v j x v 0 #$ ! ! !""" $#$#! ! if x . . . is in V (b) Prove that V V 0 . x2 2 x2 3 V , then x2 1 x2 4 x2 5 x x 0 because the left The space V is the kernel of pro jV . Since Im pro jV V , rank pro jV Thus the nullity is 5 2 3; so, the dimension of V is 3. dimV 2. $ $# 2 c. A vector in the kernel of A whose length is 3. Label the following statements as being true or false. Statements Label The orthogonal projection of n (n 3) T to a 3 dimensional subspace has rank 3. The matrices T B and T C for two different basis B and C T for the same linear transformation T of n are similar. If Im T V for a linear transformation T : V V , T is invertible. F There is a 2 2 matrix A such that Ker A Im A . T Any invertible matrix is similar to the identity matrix. F For any given two basis v 1 v 2 and u 1 u 2 of 2 , there exists a unique linear transformation T T such that T v 1 u 1 and T v 2 u 2. There is a 6 3 matrix whose image consists of all of 6 . F Two isomorphic linear spaces could have different dimension. F If x y x y , then x is perpendicular to y . F If A is any n n matrix, then Ker A Ker A2 . F If an n n matrix A has rank n, T n we have Im A For any n n matrix A, rankA5 rankA9 . F 1 is similar to B 1 . If two matrices A and B are similar, then A T If u v 0 for two nonzero vectors u and v in n , T they span two dimensional subspace of n . If B u v is a basis of 2 with u v 0, T 0 0 for the line L spanned by v . then pro jL B 01 4. Which of the following four sets are subspace of the linear space of polynomials P? Explain briefly your reason. pt : 1 0 p t dt p t : p 100 p1 1 No No zero vector Subset p t : degree 3 pt :p 0 0 yes or no reason No No zero vector No No negative vector The function p t p 100 p1 Yes is linear, and the set is the kernel of it 4 ! 4 ' ! ! 4 either (t 1) or 4 4 1 2 1 1 2 1 (t 1) has length 6. 4 By the row operation above, the kernel is given by 6.
t 2t t for all scalars t. Then ...
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 Fall '08
 lee
 Linear Algebra, Algebra

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