hw7solutions

hw7solutions - Math 215 HW #7 Solutions 1. Problem 3.3.8....

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 215 HW #7 Solutions 1. Problem 3.3.8. If P is the projection matrix onto a k -dimensional 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 . 2. Problem 3.3.12. If V is the subspace spanned by (1 , 1 , 0 , 1) and (0 , 0 , 1 , 0), find (a) a basis for the orthogonal complement V . Answer: Consider the matrix A = ± 1 1 0 1 0 0 1 0 ² . 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 = ~ 0 is equivalent to x 1 = - x 2 - x 4 x 3 = 0 . Therefore, the solutions of the homogeneous equation are of the form x 2 - 1 1 0 0 + x 4 - 1 0 0 1 , so the following is a basis for nul( A ) = V : - 1 1 0 0 , - 1 0 0 1 . (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 = 1 0 1 0 0 1 1 0 . 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Therefore, the projection matrix P onto V = col( B ) is P = B ( B T B ) - 1 B T = A T ( AA T ) - 1 A. Now, B T B = AA T = ± 1 1 0 1 0 0 1 0 ² 1 0 1 0 0 1 1 0 = ± 3 0 0 1 ² , so ( AA T ) - 1 = ± 1 3 0 0 1 ² . Therefore, P = A T ( AA T ) - 1 A = 1 0 1 0 0 1 1 0 ± 1 3 0 0 1 ²± 1 1 0 1 0 0 1 0 ² = 1 0 1 0 0 1 1 0 ± 1 3 1 3 0 1 3 0 0 1 0 ² = 1 3 1 3 0 1 3 1 3 1 3 0 1 3 0 0 1 0 1 3 1 3 0 1 3 (c) the vector in V closest to the vector ~ b = (0 , 1 , 0 , - 1) 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 double- check this since the projection of ~ b onto V is P ~ b = 1 3 1 3 0 1 3 1 3 1 3 0 1 3 0 0 1 0 1 3 1 3 0 1 3 0 1 0 - 1 = 0 0 0 0 . 3. Problem 3.3.22. Find the best line C + Dt to fit b = 4 , 2 , - 1 , 0 , 0 at times t = - 2 , - 1 , 0 , 1 , 2. Answer: If the above data points actually lay on a straight line C + Dt , we would have 1 - 2 1 - 1 1 0 1 1 1 2 ± C D ² = 4 2 - 1 0 0 . 2
Background image of page 2
Call the matrix A and the vector on the right-hand side ~ b . Of course this system is inconsistent, but we want to find b x = ± C D ² such that A b x is as close as possible to ~ b . As we’ve seen, the correct choice of b x is given by b x = ( A T A ) - 1 A T ~ b.
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This document was uploaded on 08/20/2011.

Page1 / 12

hw7solutions - Math 215 HW #7 Solutions 1. Problem 3.3.8....

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online