Math 54. Selected Solutions for Week 5
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1 Math 54. Selected Solutions for Week 5 Section 4. (Page 94) 8. Consider the following two systems of equations: 5x + x 3x 3 = 5x + x 3x 3 = 9x + x + 5x 3 = 4x + x 6x 3 = 9 9x + x + 5x 3 = 5 4x + x 6x 3 = 45 It can be shown that the first system has a solution. Use this fact and the theory from this section to explain why the second system must also have a solution. (Make no row operations.) Let A be the (common) coefficient matrix of the systems, and let b = (,, 9). If x is a solution of the first system, then A x = b, so A(5 x) = 5(A x) = 5 b, and therefore 5 x is a solution of the second system. Wait a minute. This doesn t use methods from this section. Instead, let A and b be as before. If the first system has a solution, then b lies in Col A. Since Col A is a linear subspace, 5 b also lies in Col A, and therefore the second system has a solution, too. 34. (Calculus required) Define T : C[, ] C[, ] as follows. For f in C[, ], let T ( f) be the antiderivative F of f such that F () =. Show that T is a linear transformation, and describe the kernel of T. (See the notation in Exercise of Section 4..) From calculus, we see that T ( f) = f(t) dt (strictly speaking, it is the function from [, ] to R whose value at x [, ] is the above integral). The first thing we need to check is that T ( f) C[, ] for all f C[, ]. This is true because the above antiderivative is defined for all x [, ], and is continuous on [, ] (by the Fundamental Theorem of Calculus). To check that T is a linear transformation, we check addition and scalar multiplication: T ( f + g) = ( f(t) + g(t)) dt = f(t) dt + and T (cf) = cf(t) dt = c f(t) dt = ct ( f). Therefore T is a linear transformation. g(t) dt = T ( f) + T ( g)
2 Section 4.3 (Page ) 6. Determine whether the set, is a basis for R 3. If the set is not a basis, determine whether it is linearly independent and whether it spans R 3. Justify your answers. This set is linearly independent because it has two elements and neither is a scalar multiple of the other. It does not span R 3, though. This is because the matrix with nonzero determinant 4 has linearly independent columns (by the Invertible Matrix Theorem). Therefore the first two columns are not a maximal linearly independent set, so they cannot be a basis of R 3 (see the second paragraph of Two Views of a Basis on page ). 4. Assume that A is row equivalent to B. Find bases for Nul A and Col A A =, B = The homogeneous linear system has free variables x and x 4. We have x 5 = ; 3x 3 = 6x 4 3x 5 ; and x = x x 4 5x 5. Therefore the solution space in parametric vector form is + x 4, x so a basis for Nul A is the two vectors in the above expression.
3 3 A basis for Col A is the pivot columns of A ; namely,, 3 3 3, Find a basis for the space spanned by the given vectors v,..., v 5 : 3,,, , 6 9 Row reduce the matrix whose columns are the given vectors: The first, second, fourth, and fifth columns are the pivot columns, so v, v, v 4 v 5 form a basis for Span{ v, v, v 3, v 4, v 5 }. and 6. In the vector space of all real-valued functions, find a basis for the subspace spanned by {sin t, sin t, sin t cos t}. We have sin t = sin t cos t, so sin t and sin t cos t are constant (scalar) multiples of each other. So, a basis will be {sin t, sin t} or {sin t, sin t cos t} (in each case, neither element of the set is a constant multiple of the other, so they are linearly independent). 34. Consider the polynomials p (t) = + t, p (t) = t, and p 3 (t) = (for all t ). By inspection, write a linear dependence relation among p, p, and p 3. Then find a basis for Span{ p, p, p 3 }. p + p p 3 =. We can eliminate any one of the three. The remaining two will be linearly independent since neither is a constant multiple of the other, so any two of the three will be a basis for Span{ p, p, p 3 }.
4 4 Section 4.4 (Page ). Use an inverse matrix to find [ x] B for {[ ] [ ]} 3 B =,, x = 5 [ ] 5 [ ] 3 Let A = P B =. Then det A =, so 5 [ ] [ ] A = =. Since A[ x] B = x, we have [ [ x] B = A 5 3 x = ] [ ] = 5 [ ] 5 3. The set B = { + t, t + t, + t + t } is a basis for P. Find the coordinate vector of p(t) = + 4t + 7t relative to B. As in Practice Problem # (on page ; solution on page ), we note that the coordinates of p(t) relative to B satisfy c ( + t ) + c (t + t ) + c 3 ( + t + t ) = + 4t + 7t. Simplifying the left-hand side gives This gives the linear system (c + c 3 ) + (c + c 3 )t + (c + c + c 3 )t = + 4t + 7t. c c = 4 c 3 7 Row reduce the augmented matrix of the system: Therefore [ p(t) ] B = 6..
5 8. Let B = { b,..., b n } be a basis for a vector space V. Explain why the B-coordinate vectors of b,..., b n are the columns e,..., e n of the n n identity matrix. For each i =,..., n, we have b i = b + + b i + b i + b i+ + + b n, so 5 since the is in the i th [ b i ] B = (,...,,,,..., ) = e i coordinate.. Suppose { v,..., v 4 } is a linearly dependent spanning set for a vector space V. Show that each w in V can be expressed in more than one way as a linear combination of v,..., v 4. [Hint: Let w = k v + + k 4 v 4 be an arbitrary vector in V. Use the linear dependence of { v,..., v 4 } to produce another representation of w as a linear combination of v,..., v 4.] Since v,..., v 4 span V, given any w V we can write w = k v + + k 4 v 4 for some k,..., k 4 R. Since v,..., v 4 are linearly dependent, they have a linear dependence relation c v + + c 4 v 4 =. Adding this to the above equation, we get w = w + = k v + + k 4 v 4 + c v + + c 4 v 4 = (k + c ) v + + (k 4 + c 4 ) v 4. This is a different linear combination from k v + + k 4 v 4, because c i for some i, so k i k i + c i for that value of i. 4. Show that the coordinate mapping is onto R n. That is, given any y in R n, with entries y,..., y n, produce u in V such that [ u] B = y. Given y = (y,..., y n ) R n, let u = y b + + y n bn. Then (by definition of coordinate vector), we have [ u] B = y. y n and so the coordinate mapping maps u to y. = y, 8. Use coordinate vectors to test the linear independence of the set of polynomials Explain your work. t t 3, t + t 3, + t t. As in Example 6, these vectors give a matrix A, for which the augmented matrix [ A ] for the system A x = is row reduced as follows: This system has no free variables, hence no nontrivial solutions, so the set is linearly independent.
6 6 Section 4.5 (Page 7) 5. For the subspace p q p + 5r q + r 3p + 6r : p, q, r in R, (a) find a basis for the subspace, and (b) state the dimension. In parametric vector form, this subspace equals p + q r 6 If we row reduce the matrix whose columns are the vectors occurring above, we find: / 6 6 7/ We can stop here, since it is clear that all columns are pivot columns, and therefore the vectors are linearly independent. Thus a basis for the subspace is 3 and the dimension of the space is For the subspace,, 5, 6 {(a, b, c) : a 3b + c =, b c =, b c = }, (a) find a basis for the subspace, and (b) state the dimension. This subspace is the solution set of the linear system a 3b + c = b c = b c =,
7 7 which is the null space of the matrix 3, which row reduces as follows: Since all columns are pivot columns, there are no free variables, and therefore the subspace is the trivial subspace. Therefore a basis for the subspace is and the dimension of the subspace is.,. Find the dimension of the subspace spanned by, 3,, 5 The subspace spanned by these vectors is the column space of the matrix whose columns are the given vectors, and we row reduce it as follows: This matrix has three pivot columns, and so the subspace has dimension Determine the dimensions of Nul A and Col A for the matrix This matrix is already in row reduced form, so we can read off the answers directly from the matrix. There are four pivot columns, so dim(col A) = 4. There are three free variables, so dim(nul A) = 3.
8 8 6. Let H be an n-dimensional subspace of an n-dimensional vector space V. Show that H = V. The subspace H has a basis with n vectors. These are n linearly independent vectors in V, so by the Basis Theorem, they constitute a basis for V as well. So H = V, since both H and V are equal to the set spanned by that basis. 3. Let V and W be finite-dimensional vector spaces, and let T : V W be a linear transformation. Let H be a nonzero subspace of V, and let T (H) be the set of images of vectors in H. Then T (H) is a subspace of W, by Exercise 35 in Section 4.. Prove that dim T (H) dim H. Let B = { b,..., b n } be a basis for H. Then the set T (B) = {T ( b ),..., T ( b n )} spans T (H). This is true because, for any y T (H), we can write y = T ( x) for some x H. Write x = c b + + c n bn ; then y = T ( x) = T (c b + + c n bn ) = c T ( b ) + + c n T ( b n ), and this lies in Span{T ( b ),..., T ( b n )}. Since all of T ( b ),..., T ( b n ) lie in T (H), we have that these vectors span T (H). By the Spanning Set Theorem, some subset of {T ( b ),..., T ( b n )} is a basis for T (H). The set {T ( b ),..., T ( b n )} has n elements, so the basis composed of a subset of it has at most n elements. Therefore T (H) has dimension at most n.
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