Midterm #2 Solutions

Transcription

1 Naneh Apkarian Math F Winter Midterm # Solutions Here is a solution key for the second midterm. The solutions presented here are more complete and thorough than your responses needed to be - in order to help you better understand the answers. You have two scores on your exam. The raw score, and the adjusted one. The scores were adjusted by the addition of 5 points. unadjusted average: 8/ adjusted average: /. Find the inverse of the following matrix: A = 4 8 Setup the augmented matrix: [ A I And row reduce until the left side is the identity. Then you will have [ I A ] ] Final solution: 9 7 A = 4. (a) Let W = {[ ] } x : x and y. y Determine whether or not W is a subspace of R. (b) Let U and W be two subspaces of a vector space V. Prove that the intersection of U and W, written U W, is also a subspace of V. Find an example in R which shows that the union U W is not, in general, a subspace.

2 (a) Since W is a subset of R, there are only three criteria we need concern ourselves with: [ ] (a) Is in W? (b) Is W closed under addition? i.e. if u, v W, is (u + v) W? (c) Is W closed under scalar multiplication? i.e. if u W, is cu W for all c R? If all three are true, then W is indeed a subspace of R. If any one of them is false, then W is not a subspace of R. As it happens, the third condition is violated. Here is an example of why: [ ], and u = W But R, so what about ( )u? ( )u = [ ] [ = / W. ] Thus W is not a subspace. (b) This problem was in your homework. To refresh your memory: intersection of sets: U W = {v : v U AND v W} union of sets: U W = {v : v U OR v W} As in part (a), we need to check three conditions. Both U, W are subspaces of V, which tells us that U and W, which means U W. If u, w U W, then by definition u, w U and u, w W. Since U, W are subspaces, they are closed under addition - meaning u + w U and u + w W, implying u + w U W. Any single element u U W is in U and in W, so that cu U and cu W for any scalar c, meaning of course that cu U W. Thus the [ intersection ] U W is a subspace [ ] of V. x For a counterexample, let U = { : x R} and W = { : y R}. Then y u = [ ] U U W w = [ ] W U W But u + w = [ ] / U or W (u + w) / U W.. Let W = span{v, v, v, v 4, v 5 }. Find a subset of of vectors to be a basis of W. Row reduce the augmented matrix [v v v v 4 v 5 ] to determine which columns are linearly independent. There is more than one correct answer here, but you needed three linearly independent vectors. One answer is {v, v, v }.

3 4. Let A be a 4 5 matrix and let B denote its echelon form. (a) Find bases for Col(A) and Row(A) using the labeling given. (b) Find Rank(A). (c) Find a basis for Null(A). (a) We can clearly see that B has three pivots, in columns r, r, r 4. By theorem, the corresponding columns of A form a basis for the columnspace, that is {r, r, r } is a basis for Col(A). Those pivots are in rows c, c, c. Similarly, the corresponding rows of A form a basis for the rowspace. That is, {c, c, c } is a basis for Row(A). (b) There are a few ways to see this. rank(a) = dim(col(a)) = # vectors in basis for Col(A) = rank(a) = dim(col(a)) = dim(row(a)) = # vectors in basis for Row(A) = The answer was. rank(a) = dim(col(a)) = # pivot columns of A = (c) To find a basis for Null(A), we must first establish Null(A) = {x R 4 : Ax = }, and discover what it looks like. Since B is already in reduced echelon form, and we are concerned with the homogenous equation, we may jump straight in. A solution x to Ax = is of the form x x x = x x 4 x 5 with x + x + x 5 = x x + x 5 = x free x 4 5x 5 = x 5 free x x 5 x x 5 x = x = 5x 5 x + x 5 5 x 5 Thus, 5 is a basis for Null(A). 5. (v.) The set B = { + t, + t, t + t } is a basis for P. Find the B-coordinate vector of p(t) = 6 + t t.

4 The coordinate vector of p(t) with respect to the basis B is the vector c [p(t)] B = c such that p(t) = c ( + t) + c ( + t ) + c (t + t ). c In our case, that means 6 + t t = c ( + t) + c ( + t ) + c (t + t ) There are a few ways of seeing this, but the net result must be that c + c = 6 c + c = c + c =. This can be rewritten c 6 c = c Whichever way you attack this problem, the correct answer is 5 [p(t)] B = 5. (v.) The set B = { + t, + t, t + t } is a basis for P. Find the B-coordinate vector of p(t) = 6 t + t. The coordinate vector of p(t) with respect to the basis B is the vector c [p(t)] B = c such that p(t) = c ( + t) + c ( + t ) + c (t + t ). c In our case, that means 6 t + t = c ( + t) + c ( + t ) + c (t + t ) There are a few ways of seeing this, but the net result must be that c + c = 6 c + c = c + c =. This can be rewritten c 6 c = c Whichever way you attack this problem, the correct answer is [p(t)] B = 4 4

5 6. Define a linear transformation T : P R by p() T(p(t)) = p() p() Find a basis for the kernel of T, and describe the range of T. This is almost identical to a homework problem we had. A polynomial p P has the form p(t) = a + a t + a t + a t where the coefficients a i are real numbers. Thus p() = a + a () + a () + a (t) = a. We see that an equivalent definition of the transformation T would be a T(p(t)) = a = a (since a is a scalar). a Now we can begin to discuss the kernel and the range of T. kernel(t) = {p P : T(p) = } = a + a t + a t + a t : a = Clearly we see that a must be zero for a =. Thus kernel(t) = {a t + a t + a t : a i R}. A basis of this set would be {t, t, t }, as these are three linearly independent polynomials and they span the kernel. As for the range, range(t) = {v R : T(p) = v for some p P } p() = p() : p P p() = a : a R = span Since the range of T is the span of a single vector, it is actually a line in R. 5