Midterm 2 Version B

Midterm 2 Version B - Answer Key to Second Midterm...

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Unformatted text preview: Answer Key to Second Midterm Examination, Version 2 1. Compute the following numbers and vectors (a correct answer without step by step description of computation: 10 points deduction each). Let A a. The rank of A. By row operation, we get 1 4 7 10 2 5 8 11 3 6 9 12 1 0 0 0 2 1 1 1 3 2 2 2 1 0 0 0 0 1 1 2 0 0 0 0 Thus rankA 2. b. The nullity of A. The sum of the nullity and the rank is the dimension of the source; so, The nullity is 1. 2. Label the following statements as being true or false. Statements Label 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 5 9. For any n n matrix A, rankA rankA F If two matrices A and B are similar, then A 1 is similar to B 1 . T If u v 0 for two non-zero 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 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 1 # 0 ) ' # 0 ) ' # & # # !" 6 # 5 4 2 3 0 ) # % # \$ 2 1 2 1 either (t 1) or (t 1 2 1 1 2 1 1) has length 6. By the row operation above, the kernel is given by c. A vector in the kernel of A whose length is 6. t 2t t 1 2 3 0 3 6 0 6 12 0 9 18 1 '( & 1 4 7 10 2 5 8 11 3 6 9 12 . for all scalars t. Then 2 3. Which of the following four sets are subspace of the linear space of polynomials P? Explain briefly your reason. Subset yes or no reason The function p t p 100 p1 is linear, and the set is the kernel of it No zero vector No zero vector No negative vector 4. 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 . 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. (b) Prove that V x1 x2 x5 dimV V 0 . 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 vector x in 5 , let v v1 x v1 v 2 x v 2 and w u1 x u1 u2 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. We compute for j 1 2, x v v x v j x v j 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. ) x v j v 1 v v 1 v j v 2 v v 0 & & ' & & & & & j 2 v j 0 & & & \$ & & 0 ) 0 & ) ' ' 0 ' & & & & ) # &&& ) ) ) if x . . . is in V V , then x2 1 x2 2 x2 3 x2 4 x2 5 x x 0 because the left & # 0 ' 0 0 p t : degree 3 pt :p 0 0 pt : p t dt 1 0 1 No No No 0 p t : p 100 p1 Yes ' ' ' 2. ...
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This test prep was uploaded on 04/07/2008 for the course MATH 33a taught by Professor Lee during the Fall '08 term at UCLA.

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