Linear Algebra Quiz 4. Problem 1 (Linear Transformations): 4 POINTS Show all Work! Consider the tranformation T : R 3 R 3 given by:

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1 Page 1 This is a 60 min Quiz. Please make sure you put your name on the top right hand corner of each sheet. Remember the Honors Code will be enforced! You may use your book. NO HELP FROM ANYONE. Problem 1 (Linear Transformations): 4 POINTS Show all Work! Consider the tranformation T : R R given by: T (x 1, x 2, x ) = (2x 1 + x 2 + x, x 1 + 2x 2 + x, x 1 + 2x ) (a) Show the transformation is linear. (b) Write down the standard matrix A of this linear transformation (c) Is the transformation one to one? Justify your answer using (b). (Hint think uniqueness for every b R ) (d) Is the transformation onto? (meaning every b R can be represented as a linear combination of the columns of A) Justify your answer using (b). (e) Is the transformation invertible? Justify your answer. a) Let x, y R, Show T (x + y) = T (x) + T (y) and T (cx) = ct (x) T (x + y) = T (x 1 + y 1, x 2 + y 2, x + y ) = (2(x 1 + y 1 ) + (x 2 + y 2 ) + (x + y ), (x 1 + y 1 ) +2(x 2 + y 2 ) + (x + y ), (x 1 + y 1 ) + 2(x + y )) = (2x 1 + x 2 + x, x 1 + 2x 2 + x, x 1 + 2x ) + (2y 1 + y 2 + y, y 1 + 2y 2 + y, y 1 + 2y ) = T (x) + T (y) T (cx) = T (cx 1, cx 2, cx ) = (2cx 1 + cx 2 + cx, cx 1 + 2cx 2 + cx, cx 1 + 2cx ) = c(2x 1 + x 2 + x, x 1 + 2x 2 + x, x 1 + 2x ) = ct (x) b) Plug in e 1, e 2 and e into T (x) and put output in the columns of A A = c) Reduce to find a pivot in every column hence one to one. d) Note from c) there is a pivot in every row hence onto. e) Since T (x) is onto and 1 1 it is invertible.

2 Page 2 Problem 2(Invertibility): 4 POINTS Show all Work! (a) Prove: If A I is an invertible matrix then ((A I) T ) 1 = ((A I) 1 ) T definition of inverse. Start from ((A I) T )((A I) 1 ) T = ((A I) 1 (A I)) T = I T = I ((A I) 1 ) T ((A I) T ) = ((A I)(A I) 1 ) T = I T = I (b) Determine for what values of k the matrix A = inverse in that case. Set up the augmented matrix [A I] and row reduce [ ] [ K k k K [ 2 1 k ] 1 6 1k 2 6 1k is invertible and find the So when 6 1K = 0 A is not invertible. Otherwise A 1 is on the right above ]

3 Page Problem (LU Factorization): 4 POINTS Show all Work! Given A = b = (a) Find the LU Factorization of the given matrix A. (Hint: No row interchanges) Row reduce: (R2 R2 + (/4)R1); (R R 2R1); (R R + 2R2) U = /2 2, Apply the inverse row operations to I. L = / (b) Find the solution of Ax = b using the Factorization found in (a). Solve Ly = b for y using down substitution then solve Ux=y using up substitution / , y = 2 / /2 2 /2, x =

4 Page 4 Problem 4: (Subspaces) 4 POINTS Show all Work! In the following cases is U a subspace of V (with vector addition and scalar multiplication defined as usual) If you claim U is a subspace. PROVE IT. Otherwiseive explain why not. (a) V = n, U = {p n : p(0) = 0} (homogeneous polynomials in n ). The zero polynomial is in V. Suppose we take X, Y polynomials U then (X + Y )(0) = X(0) + Y (0) = = 0. So the sum is in U. Suppose we take a scalar c and X a polynomial in U then (cx)(0) = c(x(0)) = c 0 = 0 So a scalar multiple of a vecter in U stays in U. Hence U is a subspace of V. (b) V = C[0, 1] = all continuous functions on [0,1] U = C 1 [0, 1] = all differentiable functions on [0,1] The zero function 0(t) (ie zero vecter) is differentiable so the zero vector is in U Suppose we take two differentiable functions in U. The sum is still differentiable. Suppose we take a scalar multiple of a differentiable function in U. It is still differentiable.. Hence U is a subspace of V. (c) V = n, U = n 1. The zero vector of Not a subspace. n is not in n 1

5 Page 5 Problem 5: (Basis) 4 POINTS Show all Work! Let B = {1, cos(t), cos 2 (t)} and G = {1, cos(t), cos(2t)}. identity: Assume the following trig cos(2t) = 1 + 2cos 2 (t) Let H be the subspace of functions spanned by the functions in B a) Show B is a basis for H. Hint: solve a system of equations with t = 0, π/2, π Need to show c c 2 cos(t) + c cos 2 (t) = 0 implies c 1 = c 2 = c = 0 Letting t = 0, π/2, π we get the system: Reducing we get that there is a pivot in every row so the system has only the trivial solution. so c 1 = c 2 = c = 0 b) Write the B-coordinate vectors for the vectors in G. Use them to show that G is a linearly independent set in H. We see g 1 = 1 b b b. So form the vector of coefficents to get the [g 1 ] B vecter below, and similarly for [g 2 ] B [g 1 ] B = For [g ] B we used the identity given., [g 2 ] B = 0 1 0, [g ] B = c) Explain why G is a basis of H Forming a matrix for the vectors in (b) we can row reduce to find out that the vectors are independent. Also the functions of G trivially have the same span as the functions of B. Therefore G is a basis of H.

6 Page 6 Problem 6: (Change of Coordinates)4 POINTS Show all Work! B = { 1 4, 2 0 5, Find the change of coordinate matrix from B to the standard basis in }