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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