MIT18_024S11_soln-pset2

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

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

± 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)

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

View Full Document
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
Ask a homework question - tutors are online