Math 353, Practice Midterm 1

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1 Math 353, Practice Midterm Name: This exam consists of 8 pages including this front page Ground Rules No calculator is allowed 2 Show your work for every problem unless otherwise stated Score Total

2 The following are true/false questions You don t have to justify your answers Just write down either T or F in the table below A, B, C, X, b are always matrices here (a) Let W, W 2 be subspaces of a vector space V, then W W 2 is a subspace of V (b) Let V be a vector space of dimension n Then any set of m vectors with m < n is linearly independent (c) Let T : V W be a linear transformation of vector spaces V and W If S V is a basis of V, then T (S) spans R(T ) (d) Let β be a basis of V and T : V V a linear transformation Write A = [T ] β Then T is an isomorphism if and only if N(A) = {} (e) Let A M n n (F ) be a square matrix Then the system AX = b is consistent if and only A is invertible (a) (b) (c) (d) (e) Answer T F T T F 2

3 2 Multiple Choice, A, B, C, X, b are always matrices here: (i) Suppose that A is an m n matrix with entries in F (a) If rank(a) = m and n > m then the system Ax = has infinitely many solutions (b) If N(A) = {}, then m n (c) If rank(a) = n and n m, then Ax = b has a unique solution (d) If rank(a) = n and n m, then Ax = b has at most one solution (e) If rank(a) = m and n > m, then Ax = b has infinitely many solution The correct answer is (a) (b) (d) (e) (ii) Which of the following statement is correct (a) Any set of 4 vectors in F 4 is a basis of F 4 (b) Any set of 4 vectors in F 3 is linearly dependent (c) Any set of 2 vectors in F 3 is linearly independent (d) Any set of 5 vectors in F 4 must span F 4 (e) Any linearly independent subset of F 3 is a basis of F 3 The correct answer is (b) (iii) Which of the given subsets of R 3 is a subspace? (a) The set of all vectors of the form (a, b, c) such that 2a + b = c (b) The set of all vectors of the form (a, b, c) such that a + b < c (c) The set of all vectors of the form (a, b, c) such that 2a + b = (d) The set of unit sphere (e) The set of all vectors of the form (a, a 2, a 3 ) The correct answer is (a) (iv) Which of the following map is a linear transformation x ( ) (a) L : R 3 R 2 by L( x + y y ) = y + z (b) T : P 3 (R) R by T (f(x)) = f(x)dx (c) S : P 3 (R) R by S(f(x) = f (3x) (d) L : M 2 2 (F ) M 2 2 (F ) by L(X) = AXA where A = (e) L : R R by L(x) = e x The correct answer is (b) (c) (d) ( )

4 3 Let p = +x, p 2 = x+2x 2 x 3, p 3 = +2x+2x 2 x 3, p 4 = +2x+3x 2 +x 3 be polynomials in P 3 (R) Find a basis for Span{p, p 2, p 3, p 4 } P 3 (R) Solutions: Consider equation of vectors a p + a 2 p 2 + a 3 p 3 + a 4 p 4 = with a i being unknowns compare the coefficients of each degree, we arrive the following system of linear equations a a 2 a 3 a 4 = Use elementary operation to simplify the equation, we arrive a a 2 a 3 = a 4 Then the equation has a solution a 4 =, a =, a 2 =, a 3 = So it is clear that p, p 2, p 3 are linear dependent If we remove p 3 or equivalently set a 3 = We see that this forces a = a 2 = a 4 = So p, p 2, p 4 are linearly independent and hence a basis for Span{p, p 2, p 3, p 4 } 4

5 ( ) 4 Let T : R 2 R 2 be a linear transformation such that T ( ) ( ) 2 T = 2 (a) Find T (b) Find the standard matrix representing T (c) Find nullity of T and the rank of T ( ) (d) Given a basis β = {, } Find matrix [T ] β = and Solutions: (a) Solve equation = x ( ) + y T = 3T ( ) ( ) (b) We find = 2 as the above, we have T ( ) = ( ( ) and 2 ( ) We see x = 3, y = Then 2 ) ( ) ( ) 4 T = 2 2 ( ) ( ) ( ) = + So similarly 2, T ( ) = ( ) So the standard matrix representing T is just A = (c) Since rows in A ar just multiple to each other, row space of A only has linearly dependent vector So rank(a) = rank(t ) = and the nullity of T is 2 rank(a) = (d) Let Q = ( 2 ) Then [T ] β = Q AQ = 5

6 5 Show that the following conditions are equivalent for a linear transformation T : V V (a) T 2 = T T = (b) R(T ) N(T ) Proof: Suppose T 2 = For any x = T (y) R(T ) with y V, we have T (x) = T (T (y)) = T 2 (y) = So x N(T ) Hence R(T ) N(T ) Conversely, suppose that R(T ) N(T ) Then for any x V, T (x) R(T ) N(T ) Hence T (T (x)) = T 2 (x) = for any x V That is T 2 = 6

7 6 Let V, W be vector spaces over field F Let v,, v n V be linearly independent vectors (a) Show that given w,, w n W then there exists a linear transformation T : V W such that T (v i ) = w i for all i =,, n (b) Could we drop the assumption of linear independence for so that the above statement is still true? why or why not? (c) Is T necessarily unique, why or why not? Solutions: (a) Proof : Since v i are linearly independent, then one can always extend v i, say {v,, v n, v n+,, v m }, to be a basis of V Now extend w i to m-vectors w,, w n, w n+,, w m For example, w n+ = = w m = Now by Theorem 26, there exists a unique linear transformation T : V W so that T (v i ) = w i for all i =,, m (b) No We can not drop the assumption that v i are linearly independent For example, let V = W = R, v = v 2 =, but w = and w 2 = 2 It is not possible to define a linear transformation T : R R so that T (v i ) = w i (c) Contrary to Theorem 26, T here in general is not unique ( ) as v i may not form a basis For example, let V = W = R 2, v = and w = v We ( see ) I V ((v ) = w On the other hand, consider linear transformation x x T = We still have T (v y ) = w 7

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