MIT18_024S11_soln-pset2

MIT18_024S11_soln-pset2 - Solution for PSet 2 1. (a) First,...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
± ² ± ² ± ²± ² ± ² ± ² Solution for PSet 2 1. (a) First, note T θ (1 , 0) = (cos θ, sin θ ) and T θ (0 , 1) = (cos( θ + π/ 2) , sin( θ + 2)) = ( sin cos θ ) so the matrix is cos θ sin θ sin θ cos θ (b) There are two ways to determine this problem, but perhaps the easiest is to find T θ . In that case T θ (1 , 0) = (cos( θ ) , sin( θ )) = (cos sin θ ) and T θ (0 , 1) = (cos( 2 θ ) , sin( 2 θ )) = (sin cos θ ). So T 1 = cos θ sin θ . θ sin θ cos θ Finally, to check we note cos θ sin θ cos θ sin θ cos 2 θ + sin 2 θ 0 TT 1 = = sin θ cos θ sin θ cos θ 0 cos 2 θ + sin 2 θ This evaluates to ± ² 10 01 2. First note that T (1 , 0 , 0) = (1 , 0) = 1 · (1 , 0)+0 · (1 , 1); T (1 , 1 , 0) = (1 , 1) = 2 · (1 , 0) 1 · (1 , 1); T (1 , 1 , 1) = (1 , 1) = 2 · (1 , 0) 1 · (1 , 1) and thus the matrix for T in these bases is 1 2 2 . 0 1 1 To find the matrix for S we perform the same process: S (1 , 0 , 0) = ( 1 , 0 , 0) = 1 · (1 , 0 , · (1 , 1 , · (1 , 1 , 1). S (1 , 1 , 0) = ( 1 , 1 , 0) = 0 · (1 , 0 , 0) 1 · (1 , 1 , · (1 , 1 , 1) S (1 , 1 , 1) = ( 1 , 1 , 1)=0 · (1 , 0 , · (1 , 1 , 0) 1 · (1 , 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
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 01/18/2012 for the course MATH 18.024 taught by Professor Christinebreiner during the Spring '11 term at MIT.

Page1 / 4

MIT18_024S11_soln-pset2 - Solution for PSet 2 1. (a) First,...

This preview shows document pages 1 - 2. Sign up to view the full document.

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