1) Analogous to Problem (1) in the examples, calculate the matrix if φ = π/6
We have:
F(E1) = cos(φ) (E1) + sin(φ) E2
F(E2) = -sin(φ)(E1) + cos(φ) E2
If φ = π/6:
F(E1) = cos(π/6) (E1) + sin(π/6) E2
F(E2) = -sin(π/6)(E1) + cos(π/6) E2
And we know: cos(π/6) = [3]^½/ 2 and sin(π/6) = ½,
therefore:
F(E1) = ([3]^½/ 2) E1 + ( ½)E2
F(E2) = (- ½)E1 + ([3]^½/ 2) E2
and the matric is:
([3]^½/ 2) ...... ½)
(- ½........([3]^½/ 2)
(2) Find an orthornormal basis for the subspace of R^4 generated by the following vectors:
A = (1, 1, 0, 0)
B = (1, -1, 1, 1)
C = (-1, 0, 2, 1)
For A we have:
A/ ‖A‖ = (1, 1, 0, 0)/ [1^2 + 1^2 ]^½ = (1, 1, 0, 0)/ [2 ]^½
For B we have:
B/ ‖B‖ = (1, -1, 1, 1)/ [1^2 + (-1)^2 +1^2 + 1^2]^½
= (1, -1, 1, 1)/ [4 ]^½ = (1, -1, 1, 1)/ 2
For C we have:
C/ ‖C‖ = (-1, 0, 2, 1)/ [(-1)^2 + (2)^2 +1^2 + 1^2]^½
so: C/ ‖C‖ = (-1, 0, 2, 1)/ [6 ]^½
(3) Find an orthornormal basis for the subspace of C^3 generated by:
A = (1, -1, -i) and B = (i, 1, 2)
For A we have:
A/ ‖A‖ = (1, -1, -i)/ [1^2 + (-1)^2 + (-i)]^½
But i is a complex number, i = [-1]^½ and therefore (-i)^2 = -[-1]^2 = 1
So:
A/ ‖A‖ = (1, -1, i)/ [1^2 + (-1)^2 + 1]^½ = (1, -1, i)/ [3]^½
For B we have:
B/ ‖B‖ = (i, 1, 2)/ [(i)^2 + 1^2 + 2^2]^½
= (i, 1, 2)/ [-1 + 1 + 4]^½ = (i, 1, 2)/ [4]^½ = (i, 1, 2)/ 2
4) Let V be a finite dimensional vector space over R, with positive definite product. Prove for any elements v, w in V:
‖u + v‖^2 + ‖u - v‖^2 = 2(‖u‖^2 + ‖v‖^2)
Hint: the solution uses bra-ket notation, but since this notation confuses the interface editor, we use [ ] brackets instead. (But readers should note to substitute the 'ket' > for ] and the 'bra' for the [
Then:
‖u + v‖^2 = [u + v, v + u] = [u,u] + [u,v] + [v,u] + [v,v]
But: [u,v] + [v,u] = [v,u]* + [v,u] less than or = to 2 ‖[ v,u]‖
‖u + v‖^2 = [u - v, v - u] = [u,u] - [u, v] - [v,u] + [v,v]
Whence:
‖u + v‖^2 + ‖u - v‖^2 = 2 [u,u] + 2[v,v] + 2[v,u] - 2[v,u]
= 2 {[u,u] + [v,v]}
but:
‖u‖^2 = [u,u] and ‖v‖^2 = [v,v]
hence:
‖u + v‖^2 + ‖u - v‖^2 = 2(‖u‖^2 + ‖v‖^2)
1 comment:
Mmmmmmmm.....I got all of the problems out except #4. I went over it every which way but couldn't get it. Thanks for showing the soln.
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