Solutions to Assignment 4
.(x+y) - (x-y).(x-y) = x.x + x.y + y.x + y.y + x.x - x.y - y.x + y.y = 2| x |2 + 2| y |2
1. | x + y | + | x - y | = (x+y)
2. x - y is orthogonal to x + y
(x-y).(x+y) = 0
x.x + x.y - y.x - y.y = 0
| x |2 - | y |2 = 0
| x | = | y |
Solutions to Assignment 3
1. a. This is not a subspace, since it is not closed under addition or multiplication by a scalar. For example,
are in the set but
1
0
( 0 ) and ( 1 )
( 1 ) +( 0 ) = ( 1 ) and 2 ( 1 ) = ( 2 ) are not.
0
1
1
0
0
b. This is a subsp
Assignment 1
Due:
Thursday, January 17. Nothing accepted after Tuesday, January 24. 10% off for being late, i.e.
after the 17th.
Please work by yourself on this assignment and all the other assignments that you turn in as part
of your grade. Please see me
Formulas
Trigonometric Identities
sin(x+y) = sin x cos y + cos x sin y
cos(x+y) = cos x cos y - sin x sin y
sin x sin y = [ cos(x-y) - cos(x+y) ]
cos x cos y = [ cos(x-y) + cos(x+y) ]
sin x cos y = [ sin(x+y) + sin(x-y) ]
sin2x = [ 1 - cos(2x) ]
cos2x = [
Solutions to Assignment 2
1 4 1
3 8 0
4 9 2
1. Start out with A(0) = A =
1 0 0
1 0 0
0 0 1
0 0 1
and L(0) = I = 0 1 0 and P(0) = I = 0 1 0 . Note that P(0)A = L(0)A(0). The largest
element in absolute value in column 1 is 4 which is in row 3. Exchange row
Solutions to Assignment 9
1. a. Let be an eigenvalue of R with eigenvector v. Since R2 = I on has v = Iv = R2v = RRv = Rv = Rv = v = 2v. So
v = 2v or (2 1)v = 0. Since v 0 this means 2 1 = 0 or 2 = 1 or = 1 or = -1.
b. I is a reflection whose only eigenva
Solutions to Assignment 8
1. a. A v = AAv = Av = Av = v = v.
2
2
b. Let v be an eigenvector of A2 for the eigenvalue . Then (A2 - I)v = 0. So (A - I)(A + I)v = 0. If (A + I)v = 0, then
v is an eigenvector for A with the eigenvalue - . If (A + I)v 0, let y