Math 217 Midterm 1. Winter Solutions. Question Points Score Total: 100

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1 Math 7 Midterm Winter 4 Solutions Name: Section: Question Points Score Total:

2 Math 7 Solutions Midterm, Page of 7. Write complete, precise definitions for each of the following (italicized) terms. (a) ( points) An invertible function f : X Y. A function f is invertible if y Y, a unique x X s.t. f(x) = y. (b) ( points) The kernel of an n x m matrix. Let A be this matrix. The kernel of A is ker(a) = {x R m : Ax = }. (c) ( points) The span of vectors v,..., v m R n. span(v,..., v m ) = {x : x = c v + + c m v m, for some c,..., c m R} (d) ( points) A basis of a linear subspace V of R n. A basis of V is a set of linearly independent vectors that span V.

3 Math 7 Solutions Midterm, Page of 7. State whether each statement is True or False and justify your answer. (a) (3 points) If rref(a) = rref(b), [ then ] im(a) [ = ] im(b). FALSE. Let A = and B =. rref(a) = rref(b) = im(a) is the x-axis while im(b) is the y-axis. [ ], however (b) (3 points) If V and W are linear subspaces of R n, then V W is also a linear subspace of R n. FALSE. Let V = span(e ) and W = span(e ) in R. V and W are linear subspaces of R. If V W is also a linear subspace, it must be closed under addition. e, e V W, but e + e / V W. (c) (3 points) If A and B are n x n matrices such that ker(a) = {} and ker(b) = {}, then ker(ab) = {}. TRUE. A square matrix is invertible if and only if its kernel is {}. ker(a) = {} and ker(b) = {} implies A and B are invertible. If A, B are invertible, then their product AB is also invertible, which implies ker(ab) = {}. (d) (3 points) There exists an invertible n x n matrix with exactly n non-zero entries. FALSE. An n x n matrix A is invertible if and only if rank(a) = n. If A has exactly n non-zero entries, then it must have at least one row of s. Then rank(a) < n and it cannot be invertible. (e) (3 points) If A and B are n x n matrices such that AB = B, [ then ] A = I. FALSE. Let A be any x matrix and let B =. Then AB = B.

4 Math 7 Solutions Midterm, Page 3 of 7 3. Let T : R 5 R 3 be a linear transformation with the standard matrix A = (a) (8 points) Find a basis for ker(t ). We first find rref(a). A = 3 /3 /3 /3 /3 /3 /3 Thus the fourth and fifth columns are redundant, and we have the basis basis of ker(t ) = 3,. 3 (b) (3 points) Find a basis for im(t ). A basis for im(t ) is provided by the columns in A corresponding to the pivots in rref(a): basis of im(t ) =,,. Though we note that in this case because rank(a) = 3 = dim(r 3 ), the image is all of R 3, so any three linear independent vectors in R 3 will be a basis of the image. (c) (3 points) What is rank(a)? How is rank(a) related to dim(ker(a))? There are three vectors in our basis for the image, so rank(a) = 3 = (number of columns in A) dim(ker(a)) (by the rank-nullity theorem). (d) (3 points) Is T one-to-one? Justify your answer. No. ker(a) {}, so it is clearly not one-to-one. (e) (3 points) Is T onto? Justify your answer. Yes. dim(im(a)) = 3 = dim(r 3 ), so the image is all of the target space.

5 Math 7 Solutions Midterm, Page 4 of 7 4. Let T : R R be the rotation by 6 counterclockwise. Let S : R R be the reflection about the line y = x. (a) (9 points) Find the standard matrix of the linear transformation T S. [ ] [ ] cos(π/3) sin(π/3) / A T = = 3/ sin(π/3) cos(π/3) 3/ / ([ ]) [ ] [ ] x y S = = A y x S = [ ] [ ] [ ] / 3/ A (T S) = A T A S = 3/ = / 3/ / / 3/ (b) (6 points) Find the standard matrix of the linear transformation T 9. T 9 is rotation by (6)(9) = 54 counterclockwise, effectively rotation by 8 counterclockwise or reflection across the origin. ([ ]) x T 9 = y [ ] x y = A T 9 = [ ]

6 Math 7 Solutions Midterm, Page 5 of 7 5. ( points) Let V and W be linear subspaces of R n. Define V + W = {v + w : v V, w W }. Prove that V + W is a linear subspace of R n. Solution : In order to check that V +W is a subspace we must check three properties: () V +W, () if u, u V +W then u +u V +W, and (3) if u V and λ R then λu V +W.. Since V and W are subspaces of R n, V and W. Therefore, by construction = + V + W.. Let u, u V + W. Then there are vectors v, v V and w, w W such that u = v + w and u = v + w. Now, u + u = v + w + v + w = v + v + w + w. Since V and W are subspaces, v + v V and w + w W. This reveals that u + u indeed belongs to V + W. 3. Let u V + W and λ R. Then there exist v V and w W such that u = v + w. Since λu = λ(v + w) = λv + λw and V and W are closed under scalar multiplication, we see that λu V. Having verified that V + W has the three properties necessary to be a subspace of R n, we conclude that it is indeed a subspace.

7 Math 7 Solutions Midterm, Page 6 of 7 6. A =, v =, v =, v 3 =. 3 Let T : R 3 R 3 such that T (x) = Ax. B = (v, v, v 3 ) is a basis of R 3. (a) ( points) Find B = [T ] B. T (v ) = Av = = v = v + v + v 3 = [T (v )] B = T (v ) = Av = = v + v 3 = v + v + v 3 = [T (v )] B = T (v 3 ) = Av 3 = = v + v 3 = v v + v 3 = [T (v 3 )] B = Using the column-by-column formula for the B-matrix, [T ] B = [ [T (v )] B [T (v )] B [T (v 3 )] B ] =. (b) (6 points) Prove that [T ] B = B. The standard matrix of T = T T is A. Let S = [v v v 3 ]. Then [T ] B = S A S and B = [T ] B = S AS. B = (S AS)(S AS) = S A(SS )AS = S A S = [T ] B

8 Math 7 Solutions Midterm, Page 7 of 7 7. (6 points) Let T : R 3 R 3 be the invertible linear transformation such that T =, T =, T =. Find the standard matrix for T. Noting that the given data about T tells us the output vectors of T (e j ), we can read the matrix A from the input vectors specified: A =. 8. (8 points) Prove that if A and B are n x n matrices and AB is invertible, then A and B are invertible. There are many possible solutions. For example: (Proof ) If C = AB is invertible, then C exists. Thus CC = ABC = A(BC ) = I n Thus (thm: if AB = I, A and B are invertible) A is invertible with inverse BC. Similarly, C C = C AB = (C A)B = I n, so B is invertible. (Proof ) If AB is invertible, then ker(ab) = {} (thm: invertibility kernel = {}). Then ker(b) = {x : Bx = }, so for any vector x in ker(b) we have ABx = A(Bx) = A =, so x ker(ab) and thus ker(b) ker(ab). But ker(ab) = {}, so ker(b) = {} and B is invertible. Note that therefore B is also one-to-one (thm: invertibility transformation is one-to-one). Then for any x ker(a) we have Ax = = ABB x = AB(B x). Thus B x = (because the kernel of AB is only the zero vector), and because B is one-to-one, x =. Thus the kernel of A is also {}, and it, too, is invertible. (Proof 3) Letting the columns of B be b,..., b n, AB = [Ab Ab n ]. Thus the vectors Ab,... Ab n are linearly independent (thm: invertibility column vectors are linearly independent), and so the vectors b,..., b n are linearly independent (proved in hw), and B is invertible (thm: columns are independent matrix is invertible). Then A = (AB)B is a product of invertible matrices, and is thus invertible (thm: a product of invertible matrices is invertible).