Math 215 HW #7 Solutions


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1 Math 5 HW #7 Solutions Problem 8 If P is the projection matrix onto a kdimensional subspace S of the whole space R n, what is the column space of P and what is its rank? Answer: The column space of P is S To see this, notice that, if x R n, then P x S since P projects x to S Therefore, col(p ) S On the other hand, if b S, then P b b, so S col(p ) Since containment goes both ways, we see that col(p ) S Therefore, since the rank of P is equal to the dimension of col(p ) S and since S is k dimensional, we see that the rank of P is k Problem If V is the subspace spanned by (,,, ) and (,,, ), find (a) a basis for the orthogonal complement V Answer: Consider the matrix [ A By construction, the row space of A is equal to V Therefore, since the nullspace of any matrix is the orthogonal complement of the row space, it must be the case that V nul(a) The matrix A is already in reduced echelon form, so we can see that the homogeneous equation A x is equivalent to x x x 4 x Therefore, the solutions of the homogeneous equation are of the form x + x 4, so the following is a basis for nul(a) V :, (b) the projection matrix P onto V Answer: From part (a), we have that V is the row space of A or, equivalently, V is the column space of B A T
2 Therefore, the projection matrix P onto V col(b) is P B(B T B) B T A T (AA T ) A Now, so Therefore, B T B AA T [ (AA T ) [ [, P A T (AA T ) A [ [ [ (c) the vector in V closest to the vector b (,,, ) in V Answer: The closest vector to b in V will necessarily be the projection of b onto V Since b is perpendicular to V, we know this will be the zero vector We can also doublecheck this since the projection of b onto V is P b Problem Find the best line C + Dt to fit b 4,,,, at times t,,,, Answer: If the above data points actually lay on a straight line C + Dt, we would have 4 [ C D
3 Call the matrix A and the[ vector on the righthand side b Of course this system is inconsistent, C but we want to find x such that A x is as close as possible to D b As we ve seen, the correct choice of x is given by x (A T A) A T b To compute this, first note that A T A [ [ 5 Therefore, and so (A T A) [ 5 x (A T A) A T b [ 5 [ 5 [ 4 5 [ [ 5 8 Therefore, the bestfit line for the data is 4 5 t Here are the data points and the bestfit line on the same graph:
4 4 Problem 4 Find the best straightline fit to the following measurements, and sketch your solution: y at t, y at t, y at t, y 5 at t Answer: As in Problem, if the data actually lay on a straight line y C + Dt, we would have [ C D 5 Again, this system is not solvable, but, if A is the matrix and b is the vector on the righthand side, then we want to find x such that A x is as close as possible to b This will happen when x (A T A) A T b Now, A T A [ [ 4 6 To find (A T A), we want to perform row operations on the augmented matrix [ 4 6 so that the identity matrix appears on the left To that end, scale the first row by 4 and subtract times the result from row : [ / /4 5 / Now, scale row by 5 and subtract half the result from row : [ / / / /5 Therefore, (A T A) [ / / / /5 4
5 and so x (A T A) A T b [ / / / /5 [ / / / /5 [ / /5 Therefore, the bestfit line for the data is [ [ 6 5 y 5 t Here s a plot of both the data and the bestfit line: Problem 5 Suppose that instead of a straight line, we fit the data in Problem 4 (ie # above) by a parabola y C + Dt + Et In the inconsistent system A x b that comes from the four measurements, what are the coefficient matrix A, the unknown vector x, and the data vector b? For extra credit, actually determine the bestfit parabola Answer: Since the data hasn t changed, the data vector b will be the same as in the previous problem If the data were to lie on a parabola C + Dt + Et, then we would have that C D, E 4 5 so A is the matrix above and x is the vector next to A on the lefthand side To actually determine the bestfit parabola, we just need to find x such that A x is as close as possible to b This will be the vector x (A T A) A T b 5
6 Now, A T A To find (A T A), we want to use row operations to convert the lefthand side of this augmented matrix to I: First, scale row by 4 row : and subtract twice the result from row and six times the result from / / /4 5 5 / 5 9 / Next, subtract row from row, scale row by 5 and subtract half the result from row : / / / /5 4 Finally, scale row by 4 and subtract the result from rows and : / / /4 / 9/ /4 /4 /4 /4 Therefore, and so (A T A) / / /4 / 9/ /4 /4 /4 /4 x (A T A) A T b / / /4 / 9/ /4 /4 /4 /4 / / /4 / 9/ /4 /4 /4 /4 / /
7 Thus, the bestfit parabola is y 5 t + t 5 t, which is the same as the bestfit line! 6 Problem 44 If Q and Q are orthogonal matrices, so that Q T Q I, show that Q Q is also orthogonal If Q is rotation through θ and Q is rotation through φ, what is Q Q? Can you find the trigonometric identities for sin(θ + φ) and cos(θ + φ) in the matrix multiplication Q Q? Answer: Note that (Q Q ) T (Q Q ) Q T Q T Q Q Q T IQ Q T Q I, since both Q and Q are orthogonal matrices Therefore, the columns of Q Q are orthonormal Moreover, since both Q and Q are square and must be the same size for Q Q to make sense, it must be the case that Q Q is square Therefore, since Q Q is square and has orthonormal columns, it is an orthogonal matrix If Q is rotation through and angle θ, then, as we ve seen, [ cos θ sin θ Q sin θ cos θ Likewise, if Q is rotation through and angle φ, then [ cos φ sin φ Q sin φ cos φ With these choices of Q and Q, if x is any vector in the plane R, we see that Q Q x Q (Q x), meaning that x is first rotated by an angle φ, then the result is rotated by an angle θ Of course, this is the same as rotating x by an angle θ + φ, so Q Q is precisely the matrix of the transformation which rotates the plane through an angle of θ + φ On the one hand, we know that [ [ [ cos θ sin θ cos φ sin φ cos θ cos φ sin θ sin φ cos θ sin φ + sin θ cos φ Q Q sin θ cos θ sin φ cos φ sin θ cos φ cos θ sin φ sin θ sin φ + cos θ cos φ On the other hand, the matrix which rotates the plane through an angle of θ + φ is precisely [ cos(θ + φ) sin(θ + φ) sin(θ + φ) cos(θ + φ) Hence, it must be the case that [ [ cos(θ + φ) sin(θ + φ) cos θ cos φ sin θ sin φ cos θ sin φ + sin θ cos φ sin(θ + φ) cos(θ + φ) sin θ cos φ cos θ sin φ sin θ sin φ + cos θ cos φ This implies the following trigonometric identities: cos(θ + φ) cos θ cos φ sin θ sin φ sin(θ + φ) cos θ sin φ + sin θ cos φ 7
8 7 Problem 46 Find a third column so that the matrix / / 4 Q / / 4 / / 4 is orthogonal It must be a unit vector that is orthogonal to the other columns; how much freedom does this leave? Verify that the rows automatically become orthonormal at the same time Answer: Let q / / / and q q, q and q, q then we have that / 4 / 4 / 4 q q, q a + b + c If q q, q a/ + b/ + c/ q, q a/ 4 + b/ 4 c/ 4 a b c such that q, Multiplying the second line by and the third line by 4, we get the equivalent system a + b + c a + b + c a + b c From the second line we have that b a c and so, from the third line, a b + c ( a c) + c a + 5c Thus a 5c, meaning that b a c ( 5c) c 4c Therefore a + b + c ( 5c) + (4c) + c 4c, meaning that c ±/ 4 Thus, we see that q a b c 5c 4c c ± 5/ 4 4/ 4 / 4 Therefore, there are two possible choices; one of them gives the following orthogonal matrix: Q / / 4 5/ 4 / / 4 4/ 4 / / 4 / 4 It is straightforward to check that each row has length and is perpendicular to the other rows 8
9 [ [ 4 8 Problem 4 What multiple of a should be subtracted from a to make [ 4 the result orthogonal to a? Factor into QR with orthonormal vectors in Q Answer: Let s do GramSchmidt on { a, a } First, we let v a a [ [ / / Next, w a v, a v [ 4 4/ [ / / [ 4 [ [ By construction, w is orthogonal to a, so we see that we needed to subtract times a from a to get a vector perpendicular to a Now, continuing with GramSchmidt, we get that v w w [ Therefore, if A [ a a and Q [ v v, then A QR [ / / where R [ a, v a, v a, v [ 9 Problem 48 If A QR, find a simple formula for the projection matrix P onto the column space of A Answer: If A QR, then A T A (QR) T (QR) R T Q T QR R T IR R T R, since Q is an orthogonal matrix (meaning Q T Q I) Thus, the projection matrix P onto the column space of A is given by P A(A T A) A T QR(R T R) (QR) T QRR (R T ) R T Q T QQ T (provided, of course, that R is invertible) Problem 4 (a) Find a basis for the subspace S in R 4 spanned by all solutions of x + x + x x 4 9
10 Answer: The solutions of the given equation are, equivalently, solutions of the matrix equation x [ x x, x 4 so S is the nullspace of the 4 matrix A [ Since A is already in reduced echelon form, we can read off that the solutions to the above matrix equation are the vectors of the form x + x + x 4 Therefore, a basis for nul(a) S is given by,, (b) Find a basis for the orthogonal complement S Answer: Since S nul(a), it must be the case that S is the row space of A Hence, the one row of A gives a basis for S, meaning that the following is a basis for S : (c) Find b in S and b in S so that b + b b (,,, ) Answer: For any b S, we know that b is a linear combination of elements of the basis for S that we found in part (a) In other words, b a + b + c for some choice of a, b, c R Also, if b S, then b is a multiple of the basis vector for S we found in part (b) Thus, b d
11 for some d R Therefore, b a + b + c + d or, equivalently, a b c d To solve this matrix equation, we just do elimination on the augmented matrix Add row to row : Next, add row to row : Finally, subtract row from row 4: Therefore, 4d, so d Hence, 4 c + d c +, so c In turn, b + c + d b + + b + 5, meaning b Finally, a b + c + d a + + a +,
12 so a Therefore, and b + b + / / / / / / / / (Bonus Problem) Problem 44 Find the fourth Legendre polynomial It is a cubic x + ax + bx + c that is orthogonal to, x, and x over the interval x Answer: We can find the fourth Legendre polynomial in the same style as Strang finds the third Legendre polynomial on p 85: v 4 x, x, x, x x x, x x, x ( x, x x ) () Now, we just compute each of the inner products in turn: Therefore, () becomes, x, x, x x, x x, x x, x x dx dx x 4 dx 5 x dx (x 5 x ( x v 4 x /5 / x Therefore, the fourth Legendre polynomial is x 5 x ) dx ) dx 8 45 ( x ) x 5 x
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