103 MATH SAMPLE FINAL SOLUTIONS
1. Consider the data points (-1, 15), (0, 8), (1, 5), (2, 0). (a) Find the least-squares line for the data points above. Solution. Let f (x) = c0 + c1 x. Then c0 and c1 must satisfy c0 - c1 c0 c0 + c1 c0 + 2c1 This system is inconsistent. Since AT A = 4 2 2 6 4 2 2 6 Its reduced row echelon form is 1 0 0 1 Thus the least-squares line is f (x) = 47 24 - x 5 5 47/5 -24/5 AT b = 28 -10 = 15 = 8 = 5 = 0
the augmented matrix for the normal equations is 28 -10
(b) Plot the points and the least-squares approximation. Solution.
Least-Squares Line 20
15
10
5
0
-5 -2
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
(c) Find a cubic polynomial of the form f (x) = c0 + c1 x + c2 x2 + c3 x3 which passes through all four points. Solution. The coefficients must satisfy the system Ax = b where 1 -1 1 -1 15 1 0 0 0 8 A= b= 1 1 1 1 5 1 2 4 8 0 The solution is 8 -4 x= 2 -1
so f (x) = 8 - 4x + 2x2 - x3 . 2. Find an orthonormal basis for the subspace of 2 1 2 1 , , 2 3 2 3 R4 spanned by the following vectors: 0 2 . 4 6
Solution. Call these vectors v1 , v2 and v3 . Normalizing the first vector gives 1/2 1/2 v1 w1 = = 1/2 v1 1/2 Next, 1 2 -1 1 2 -1 v2 - (v2 w1 )w1 = v2 - 4w1 = - = 3 2 1 3 2 1 -1/2 -1/2 w2 = 1/2 1/2
so normalizing gives
Finally 0 3 -2 -1 2 3 -2 1 v3 - (v3 w1 )w1 - (v3 w2 )w2 = v3 - 6w1 - 4w2 = - - = 4 3 2 -1 6 3 2 1
so normalizing gives
-1/2 1/2 w3 = -1/2 1/2
3. Let V be any subspace of Rn . Let P1 denote the matrix for the orthogonal projection onto V and let P2 denote the matrix for the orthogonal projection onto V . (a) Show that P1 and P2 are symmetric. Hint: Use Fact 5.3.10. Solution. By Fact 5.3.10, P1 takes the form AAT for some matrix A. Thus P T = (AAT )T = (AT )T AT = AAT = P1 , so P1 is symmetric. Similarly P2 is symmetric. (b) Show that P1 P2 = P2 P1 = 0 (the zero matrix). Hint: Explain why the columns (and rows) of P1 are vectors in V . Where are the columns and rows of P2 ? Solution. The image of P1 must be V . But the image of P1 is the span of the columns of P1 . Thus each column of P1 is a vector in V , and since P1 is symmetric by part (a), the rows of P1 are also vectors in V . Likewise the rows and columns of P2 are vectors in V . Each entry in P1 P2 is the dot product of a row of P1 with a column of P2 , so each entry is the dot product of a vector in V with a vector in V . By definition of V , all such dot products are zero. Hence P1 P2 = 0, and similarly P2 P1 = 0. (c) Show that P1 + P2 = In . Hint: For any x Rn , consider w = x - P1 x. Where is w? Now apply P2 to both sides. Solution. Let x Rn . By Fact 5.1.6, the projection of x onto V is the unique vector v such that x - v is in V . Since v = P1 x this means that w = x - P1 x is in V . Applying P2 to both sides gives P2 w = P2 x - P2 P1 x = P2 x But since w is in V , P2 w = w, so w = P2 x. Thus x = P1 x + P2 x = (P1 + P2 )x for all x Rn . This means that P1 + P2 is the matrix for the identity transformation, i.e. P1 + P2 = In . 4. Let V = {x R4 | x1 + x2 + x3 + x4 = 0}. (a) Find a basis for V . Solution.
1 1 V = Span 1 1
=
V
1 1 = Span 1 1
(b) Find the matrix (in standard coordinates) for the orthogonal projection onto V . Solution. Since V is spanned by the unit vector 1/2 1/2 u= 1/2 1/2 the matrix for the orthogonal projection onto V is (by Fact 5.3.10) 1 1 1 1 1 1 1 1 1 P2 = uuT = 4 1 1 1 1 1 1 1 1 (c) Find the matrix (in standard coordinates) for the orthogonal projection onto V . Hint: Use Question 3c. Solution. By Question 3c, the matrix for the orthogonal projection onto V is 3 -1 -1 -1 1 -1 3 -1 -1 P1 = I4 - P2 = -1 -1 3 -1 4 -1 -1 -1 3 5. Let 1 1 0 A= 1 0 1 0 1 1
(a) Find all eigenvalues of A and a basis for the eigenspace associated with each eigenvalue. Solution. The characteristic polynomial is fA () = ( - 1)( - 2)( + 1), so the eigenvalues are = 1, 2, -1. The eigenspaces are -1 1 1 0 E2 = Span 1 E-1 = Span -2 E1 = Span 1 1 1 (b) Diagonalize A. That is, find an invertible matrix S that S -1 AS = D. Solution. -1 1 1 S = 0 1 -2 D= 1 1 1 and a diagonal matrix D such 1 0 0 0 2 0 0 0 -1
6 0. (c) Solve the system x(t + 1) = Ax(t) with initial data x0 = 0 Solution. Since 6 -1 1 1 0 = -3 0 + 2 1 + 1 -2 0 1 1 1 the solution is 1 1 -1 0 + 2(2)t 1 + (-1)t -2 x(t) = -3 1 1 1 6. Consider the plane V = {x R3 | x1 + x2 + x3 = 0}. (a) Find a basis {v1 , v2 } for V and a basis {v3 } for V . Solution. -1 -1 1 1 , 0 1 V = Span V = Span 0 1 1 (b) Let T : R3 R3 be the linear transformation defined by reflection through V . That is, T sends each vector in R3 to its mirror image on the opposite side of V . Write down the matrix B for T with respect to the basis B = {v1 , v2 , v3 } of R3 . Solution. T (v1 ) = v1 = 1v1 + 0v2 + 0v3 T (v2 ) = v2 = 0v1 + 1v2 + 0v3 T (v3 ) = -v1 = 0v1 + 0v2 + (-1)v3 Thus 1 0 0 0 B= 0 1 0 0 -1
(c) Write down the matrix A for T with respect to the standard basis for R3 . Solution. Let -1 -1 1 0 ...