linalg 2 test 1

vector space

(V,+,*) a.r V is nonempty/closed under add/scalar mult

25. Find the vector x determined by the given coordinate vector [x]B and the given basis B B= { [ -3] , [ 4 ] , [ 5 ] } , [x]B = [ -2 ] -1 0 -2 -1 5 1 3 -3

-2 [first vector], -1 [2nd vector], -3 [3rd vector] x = [ -13 ] 8 -20

spanning set thm

1. if one of vectors in S={v1..vp} is lin combo of others, vector can be removed from S w/o changing Span 2. if H=Span{v1..vp} is not 0, same subset of S is basis of H

21. Suppose R4 = Span {v1 v2 v3 v4 }. Explain why {v1 v2 v3 v4 } is a basis for R4. Let A = [v1 v2 v3 v4 ]. Note that A is a (1) matrix and its columns span (2). thus by the (3) the columns (4). Therefore the columns of A are a basis for R4 because of the (5).

1. 4x4 2. R4 3. Invertible Matrix Theorem 4. are linearly independent 5. definition of a basis

3 requirements to be a basis

1. B={b1..bp} is linear indep 2. span of {b1..bp} = subspace H 3. A is row equiv to identity matrix

10. It can be shown that a solution of the system below is x1 = 4, x2 = 3 , x3 = -3. Use this fact and the theory of null spaces and column spaces of matrices to explain why another solution is x1 = 400, x2 = 300, x3 = -300. (Observe how the solutions are related, but make no other calculations.) -6x1 -16x2 -24x3 = 0 -3x1 + 12x2 + 8x3 = 0 -12x1 + 24x2 + 8x3 =0 Let A be the coefficient matrix of the given homogeneous system of equations. The vector x = [ 4 ] 3 -3 is in the vector space (1). Next, determine the relationship between the given solution and proposed solution (1a). Notice that the proposed solution vector is (2). Since all vector spaces (3) [400 300 -300] must be in (4). the proof is complete bc if [400 300 -300] is in (5). then is a sol to ax=0

1. Nul A 1a. 100 2. scalar mult of x 3. closed under scalar mult 4. Nul A 5. Nul A

3. Determine if the given set is a subspace of P6. All polynomials of degree 6 or less, negative real #s as coefficients. the zero vector of P6 is (1) in the set because zero (2) a negative real number. the set (3) closed under addition bc sum of 2 negative numbers (4) a negative number. the set (5) closed under multiplication by scalars because the product of a scalar and a negative real number (6) a negative real number. is this a subset of P6? (7)

1. is not 2. is not 3. is 4. is 5. is not 6. is not necessarily 7. no

the long way to find an inverse

1/det A, det swap diagonals of 2x2 and change signs of others

13. Either use an appropriate theorem to show that the given set, W, is a vector space, or find a specific example to the contrary. W = [ s -2t ] 2+3s 3s+t 2s a. The set W is a subset of R4. If W were a vector space, what else would be true about it? b. Determine whether the zero vector is in W. Find values for t and s such that W = [ 0 ] 0 0 0 c. what is the true statement?

13a. set W would be a subspace of R4 13b. zero vector is not in W. There is no t and s s.t the vector eq is satisfied 13c. Since the zero vector is not in W, W is not a subspace of R4. Thus W is not a vector space

29. If B is the standard basis of the space P3 of polynomials, then let B = {1, t, t^2, t^3} Use coordinate vectors to test the linear independence of the set of polynomials below. Explain your work (-1-t)^3 , (2-t)^2, 5-1t+4t^2+t^3 29a. write coordinate vectors for above p1, p2, p3 29b. To test the linear independence of the set of polynomials, row reduce the matrix which is formed by making each coordinate vector a column of the matrix. Are the polynomials linearly independent?

29a. use coefficients as vars in vectors p1 = [ -1 ] -3 -3 -1 p2 = [ 4 ] -4 1 0 p3 = [ 5 ] -1 4 1 29b. These coordinate vectors are dependent, as p3 = -1p1 + 1p2

31. For the subspace below, (a) find a basis, and (b) state the dimension. [ 12a + 24b − 4c ] 9a − 3b − 3c − 3a + 5b + c − 3a + b + c

31a. basis is coefficients of first two cols [12 ] , [24 ] 9 -3 -3 5 -3 1 31b. dimension is 2 bc two vectors from part a

32. For the subspace below, (a) find a basis, and (b) state the dimension. [a ] b c d : a − 5b + c = 0

32a. use coefficients of b,a,c 5b -a -c [ 5] [ -1] [ 0] 1 0 0 0 1 0 0 0 1 32b. dimension is 3 bc three vectors from part a

35. V is a nonzero finitedimensional vector space, and the vectors listed belong to V. Mark each of the following statements as True or False. Justify each answer. Complete parts (a) through (c) below. a. If there exists a set { v1 , ... , vp } that spans V, then dim V <= p. Choose the correct answer b. If there exists a linearly independent set {v1...vp} n V, then dim V >=p. choose correct answer. c. If dim V p, then there exists a spanning set of p 1 vectors in V. Choose the correct answer below.

35a. True. Apply the Spanning Set Theorem to the set {v1...vp} and produce a basis for V. This basis will not have more than p elements in it, so dim V <= p. 35b. True. The span of the set {v1...vp} will be a subspace of V which can be expanded to find a basis for V. This basis will contain at least p elements, so dim V >= p 35c. True. The spanning set of p 1 can be found by taking any spanning set, which contains p vectors, for V, and adjoining the zero vector to it.

37. A linearly independent set {v1...vk} in Rn can be expanded to a basis for Rn. One way to do this is to create A = [ v1 .... vk e1 .... en] with e1...en the columns of the identity matrix; the pivot columns of A form a basis for Rn. choose correct answer a. Use the method described to extend the following vectors to a basis for R5 . Choose the correct answer below. v1 v2 v3 [7 ] [ 5 ] [ -7 ] − 6 3 6 − 9 5 5 − 2 2 2 6 -8 -6 b. Explain why the method works in general. Why are the original vectors v1...vk included in the basis found for Col A ? The original vectors are the first k columns of A. Since the set of original vectors is assumed to (1) linearly independent, these columns of A will be (2) columns and the (3) will be included in the basis. Why is Col A = Rn? since all of the columns of the n xn identity matrix are (4) of A, every vector in (5) is in (6) Since every column of A is in Rn every vector in (7) is in (8) This shows that Col A and Rn are equivalent

37a. {v1,v2,v3, e1,e2} 37b. 1. be 2. pivot 3. original set of vectors 4. columns 5. Rn 6. Col A 7. Col A 8. Rn

22. let v1, v2, v3 equal [ 1 ] [ 0 ] [ 0 ] 0 1 1 1 1 0 and let H be the set of vectors in R3 whose second and third entries are equal. Then every vector in H has a unique expansion as a linear combination of v1,v2, and v3 because the following equation is true for any s and t [ s ] = s [ 1 ] + (t-s) [ 0 ] + s [ 0 ] t 0 1 1 t 1 1 0 is the set S = {v1 v2 v3 } a basis for H? Why or why not?

The set S is not a basis for H because there are elements in Span {v1 v2 v3} that are not in H

24. Consider the polynomials p1 (t) = 4 + 5t^2 and p2 (t) = 4- 5t^2. Is {p1 p2} a linearly independent set in P3? Why or why not?

The set {p1, p2} is a linearly independent set because neither polynomial is a multiple of the other polynomial.

30. Let B be the standard basis of the space P2 f polynomials.Use coordinate vectors to test whether the following set of polynomials span P2 Justify your conclusion. − 2t + t^2, 1 + 17t − 5t^2, − 4 − 11t + t^2 , − 3 + 14t − 4t^2 Does the set of polynomials span P2?

Yes ; since the matrix whose columns are the B coordinate vectors of each polynomial has a pivot position in each row, the set of coordinate vectors spans R3. By isomorphism between R3 an P2 the set of polynomails spans P2

invertible matrix thm 6 parts

a has inverse cols of a form basis for Rn col (a) = Rn rank(a) = n nul (a) = {0} dim nul(a) = 0

14. A = [ 2 − 9 0 − 3 ] − 7 − 1 8 1 − 3 4 − 4 − 9 5 4 − 4 − 7 − 4 5 4 − 7 a. find k s.t nul A is part of subspace of Rk b. find k s.t col A is part of subspace of Rk

a. 4 cols so 4 b. 5 rows so 5

17. let a1,a2 ... a5 denote cols of matrix A and let b = [a1 a2 a4] A= 3 3 2 3 − 21 6 − 3 1 4 39 3 − 9 − 2 4 76 3 − 6 − 1 − 2 69 a.Explain why a3 and a5 are in the column space of B. b. Find a set of vectors that spans Nul A c.Let T : R5->R4 be defined by T(x ) = Ax. Explain why T is neither one-toone nor onto. Choose the correct answer below.

a. The free variable columns can always be written as a linear combination of the pivot columns. b. rref A [ -1/3 ] [ -13/3 ] -1/3 25/3 1 0 0 3 0 1 c. the reduced row echelon form of A shows that the columns of A are linearly dependent/do not span R4

16. For parts a. through f., A denotes an m x n matrix. Determine whether each statement is true or false. Justify each answer. a. A null space is a vector space. Is this statement true or false? b. The column space of an m n matrix is in Rm Is this statement true or false c.The column space of A, Col(A), is the set of all solutions of Ax=b Is this statement true or false? d. the null space of A Nul(a) is the kernel of the mapping x->Ax is this statement true or false e. The range of a linear transformation is a vector space. Is this statement true or false? f. The set of all solutions of a homogeneous linear differential equation is the kernel of a linear transformation. Is this statement true or false

a. true bc null space of an mxn matrix A is subspace of Rn b. true bc the col space of an mxn matrix A is a subspace of Rm c. false bc col(a) = {b:b=ax for some x in Rn } d. true the kernel of a lin transform T, from a vector space V to a vector space W is the set of all u in V s.t T(u) = 0. thus the kernel of fa matrix transform T(x) = Ax is the null space of A e. True, the range of a linear transformation T, from a vector space V to a vector space W, is a subspace of W. f. True, the linear transformation maps each function f to a linear combination of f and at least one of its derivatives, exactly as these appear in the homogeneous linear differential equation

linear independent

all alphas 0, trivial sol, all sols =0, no free vars

12. find explicit description of Nul A by listing vectors that span the null space A = [ 1 5 − 4 − 7 1 ] 0 1 − 8 1 0 0 0 0 0 0

aug matrix, rref, set up xs as free/basic vars, write basic in terms of free, pull out x's spanning set is [ -36] [ 12 ] [ -1 ] 8 -1 0 1 0 0 0 1 0 0 0 1

20. Assume that A is row equivalent to B. Find bases for Nul A and Col A. A= [-2 8 -2 -8 ] 2 -12 -4 4 -3 16 3 -8 B= [1 0 7 8 ] 0 4 6 4 0 0 0 0

basis for col(a) is first two cols [ -2 ], [ 8 ] 2 -12 -3 16 rref A, basis for nul(a) [ -7 ] , [ -8 ] -3/2 -1 1 0 0 1

2. Let W be the union of the 2nd and 4th quadrants in the xy plane. That is, let W = [x ] : xy less equal 0 [y ] a. If u is in W and c is any scalar, is cu in W? Why?

bc (cx)(cy) = c^2(xy) <= 0 since xy <= 0

if A (mxn matrix) is invertible then

cols of A form basis for Rn

34. Determine the dimensions of Nul A and Col A for the matrix shown below. A = [ 1 4 − 5 3 − 4 6 0 ] 0 0 0 1 5 − 1 − 5 0 0 0 0 1 − 2 − 6 0 0 0 0 0 0 0

dim of nul (a) is 4 bc 4 nonpivot/free var cols dim of col (a) is 3 bc 3 pivots

33. Find the dimension of the subspace of all vectors in R8 whose and entries are equal.

dimension is 7

thm regarding row equivalences of rows a & b

if 2 matrices row equiv then row(a) = row(b), if b in echelon form, nonzero rows of b form basis for row space of a

thm when is it that a set of vectors is ld

indexed set of 2+ vectors with v1 not equal 0, is ld iff exist vector that is lin combo of rest

subspace

made up of elements in V, including zero vector, closed under add/scalar mult

36. Let B be the basis of P3 consisting of the Hermite polynomials 1, 2t, -2+4t^2 , and -12+8t^3 and let p(t) = 13-16t^2 +8t^3 Find the coordinate vector of p relative to B

make a vector of coefficients of each 1,t,t^2, t^3 and for p(t), put in aug matrix, rref, then take last col

18. Determine whether the set [ 1 ] [ -3 ] [ -1 ] 0 1 1 3 -12 -6 is a basis for R3. If the set is not a basis, determine whether the set is linearly independent and whether the set spans R3 Which of the following describe the set? Select all that apply set is linear indep set spans R3 the set is a basis for R3 none of the above

none of the above

thm regarding pivot cols of Amxn

pivot cols form basis for col(a)

26. The crystal lattice for titanium has the hexagonal structure shown in the accompanying figure. The vectors b1 b2 b3 [ 2.6 ] [ 0.0 ] [ 0.0 ] -4.5 3.0 0.0 0.0 0.0 4.8 in R3 form a basis for the unit cell shown in the accompanying figure. One of the octahedral sites is [ -1/2 1/4 5/6 ]^T , relative to the lattice basis. Determine the coordinates of this site relative to the standard basis of R3 Determine the coordinates of this site relative to the standard basis of R3

put b1,b2,b3 into matrix, multiply by other matrix x = [ − 1.3 ] 1.5 4

27. Use an inverse matrix to find [x]B for the given x and B B { [ 11 ] , [ -2 ] } x = [ 3 ] -5 1 -4

put both vectors into one matrix A, then A^-1 * B [x]B = [ 5 ] 29

rank thm

rank(a) = dim (row (a)) = dim((col(a)) rank(a) is # of pivots in a rank(a) + dim nul(a) = n, # of cols in a

9. let A = and w = [ -6 -4 -2 ] [ -1 ] 8 6 4 2 2 0 -2 -1 determine if w is in col (a)? is w in nul(a)?

rref aug matrix, if no row is 0 0 b then consistent, if consistent is in col a. Aw does equal zero vector so yes w in nul(a)

7. The set M 2x2 of all 2x2 matrices is a vector space, under the usual operations of addition of matrices and multiplication by real scalars. Determine if the set H of all matrices of the form [ a b ] 0 d is a subspace of M 2x2

set h is subspace of M 2x2 bc h contains zero vector, closed under addition/scalar mult

8. for fixed positive integers m and n, the set M mxn of all mxn matrices is a vector space, under the usual operations of addition of matrices and multiplication by real scalars. let F be a fixed 3x2 matrix and let H be the set of all matrices A in M 2x4 with the property that FA = 0 (the zero matrix in M 3x4) determine if h is subspace of M 2x4

set h is susbspace of m 2x4 bc contains zero vector/closed under add/scalar mult

4. Determine if the given set is a subspace of Pn. Justify your answer. The set of all polynomials in Pn such that p (0) = 0

set is a subspace of Pn bc set contains zero vector/closed under addition/closed under scalar mult

col space

set of all lin combos of cols of A

Span

set of all lin combos of vectors

null space

set of all sols to homogeneous eq A* = [0]

19. Determine if the set of vectors shown to the right is a basis for R3. If the set of vectors is not a basis determine whether it is linearly independent and whether the set spans R3. [ 1 ] [ -5 ] [ 0 ] [ 0 ] -3 8 0 -3 0 0 0 8

set spans R3

11. Find an explicit description of Nul A by listing vectors that span the null space. A = [ 1 5 0 3 0 ] 0 0 1 -3 0 0 0 0 0 -1

spanning set is [ -5 ] [ -3 ] 1 0 0 3 0 1 0 0 rref aug matrix, write as sys of eqs, pull out x vars, spanning set

6. Let W be the set of all vectors of the form shown, where a and b represent arbitrary real numbers. Find a set S of vectors that spans W, or give an example or an explanation showing why W is not a vector space [ 9a − 7b ] 6b − 7c 4c − 2a 8b

spanning set is each var a b c set apart [9 0 -2 0] [-7 6 0 8 ] [0 -7 4 0]

5. Let W be the set of all vectors of the form shown, where a and b represent arbitrary real numbers. Find a set S of vectors that spans W, or give an example or an explanation showing why W is not a vector space [ 8a + 7b ] − 6 8a − 5b

w not a vector space bc zero vector/most sums/scalar mults of vectors in w are not in w bc second value is not equal to -6

closed under addition

when add vectors together, could get a vector in vector space/subspace

closed under scalar multiplication

when mult by scalar get a vector in vector space/subspace

15. Let H= Span{ v1 , v2 } and K = Span{ v3 ,v3=4 }, where v1 ,v2 ,v3 , and v4 are given below. v1 v2 v3 v4 [ 1 ] [ 5 ] [ 9 ] [ 0 ] 3 4 -1 -15 4 9 3 -9 Then H and K are subspaces of R3. In fact, H and K are planes in R3 through the origin, and they intersect in a line through 0. Find a nonzero vector w that generates that line

write as var eqs, aug matrix, rref, set up as var eqs again, rewrite as c3v3 + c4v4, take ans w= [ -2 ] 3 1

23. Let V be the vector space of functions of the form y(t) = c1 cos wt +c2 sin wt where w is a fixed constant and c1 and c2 are arbitrary (varying) constants. find a basis for v.

{cos wt, sin wt}