351Hmw9solns[1]

351Hmw9solns[1] - HW 9 solutions April 15, 2011 Section 5.1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
HW 9 solutions April 15, 2011 Section 5.1 12) || ~v + ~w || 2 = ( ~v + ~w ) . ( ~v + ~w ) = || ~v || 2 + || ~w || 2 + 2 ~v.~w ≤ || ~v || 2 + || ~w || 2 + 2 || ~v |||| ~w || ( by Cauchy Schwartz inequality) = ( || ~v || + || ~w || ) 2 Taking squareroot on both sides, we get || ~v + ~w || ≤ || ~v || + || ~w || 17) W = Span 1 2 3 4 , 5 6 7 8 If ~v W , then v 1 + 2 v 2 + 3 v 3 + 4 v 4 = 0 5 v 1 + 6 v 2 + 7 v 3 + 8 v 4 = 0 We consider the matrix ± 1 2 3 4 5 6 7 8 ² The RREF of this matrix is ± 1 0 - 1 - 2 0 1 2 3 ² Setting v 3 = s ,v 4 = t we get v 1 = s + 2 t,v 2 = - 2 s - 3 t . Hence v 1 v 2 v 3 v 4 = s + 2 t - 2 s - 3 t s t = s. 1 - 2 1 0 + t. 2 - 3 0 1 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
Hence the basis of W is 1 - 2 1 0 , 2 - 3 0 1
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 4

351Hmw9solns[1] - HW 9 solutions April 15, 2011 Section 5.1...

This preview shows document pages 1 - 3. Sign up to view the full document.

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