how to prove.

u* (vxw) = (uxv) *w

and

c(uxv) = cuxv= u x cv

thank you.

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- Apr 29th 2007, 01:29 AMtest2k6Application of Inner Product Spaces
how to prove.

**u*** (**v**x**w**) = (**u**x**v**) ***w**

and

c(**u**x**v**) = c**u**x**v**= u x c**v**

thank you. - Apr 29th 2007, 06:53 AMThePerfectHacker
- Apr 29th 2007, 12:13 PMecMathGeek
I assume * is the dot product.

**u*** (**v**x**w**) = (**u**x**v**) ***w**

Let**u**,**v**, and**w**, all be vectors of 3 dimensions:

(I'll have to stick with 3 because proving this for n dimensions is a bit difficult.)

**u**= <u1, u2, u3>

**v**= <v1, v2, v3>

**w**= <w1, w2, w3>

Then:

**u*** (**v**x**w**)

= <u1, u2, u3> * (<v1, v2, v3> x <w1, w2, w3>)

= <u1, u2, u3> * <v2*w3 - v3*w2, v3*w1 - v1*w3, v1*w3 - v3*w1>

= u1(v2*w3 - v3*w2) + u2(v3*w1 - v1*w3) + u3(v1*w2 - v2*w1)

= u1*v2*w3 - u1*v3*w2 + u2*v3*w1 - u2*v1*w3 + u3*v1*w2 - u3*v2*w1

= u2*v3*w1 - u3*v2*w1- u1*v3*w2 + u3*v1*w2 + u1*v2*w3 - u2*v1*w3

= (u2*v3 - u3*v2)w1 + (u3*v1 - u1*v3)w2 + (u1*v2 - u2*v1)w3

= (<u1, u2, u3> x <v1, v2, v3>) * <w1, w2, w3>

= (**u**x**v**) ***w**

Q.E.D. - Apr 29th 2007, 12:30 PMecMathGeek
Let

**u**and**v**be vectors of 3 dimensions:

**u**= <u1, u2, u3>

**v**= <v1, v2, v3>

c(**u**x**v**)

= c(<u1, u2, u3> x <v1, v2, v3>)

= c(<u2*v3 - u3*v2, u3*v1 - u1*v3, u1*v2 - u2*v1>)

= <c*u2*v3 - c*u3*v2, c*u3*v1 - c*u1*v3, c*u1*v2 - c*u2*v1>

From this there are two thing we have to show:

1. c(**u**x**v**) = c**u**x**v**

c(**u**x**v**) = <c*u2*v3 - c*u3*v2, c*u3*v1 - c*u1*v3, c*u1*v2 - c*u2*v1>

= <(c*u2)v3 - (c*u3)v2, (c*u3)v1 - (c*u1)v3, (c*u1)v2 - (c*u2)v1>

= <c*u1, c*u2, c*u3> x <v1, v2, v3>

= (c*<u1, u2, u3>) x <v1, v2, v3>

= c**u**x**v**

2. c(**u**x**v**) =**u**x c**v**

c(**u**x**v**) = <c*u2*v3 - c*u3*v2, c*u3*v1 - c*u1*v3, c*u1*v2 - c*u2*v1>

= <u2(c*v3) - u3(c*v2), u3(c*v1) - u1(c*v3), u1(c*v2) - u2(c*v1)>

= <u1, u2, u3> x <c*v1, c*v2, c*v3>

= <u1, u2, u3> x (c*<v1, v2, v3>)

=**u**x c**v**

NOTE: in this problem "*" is not being used as the dot product because c is a scalar constant. The symbol "*" simply represents scalar multiplication. - Apr 30th 2007, 03:35 AMtest2k6
thank you, * is the dot product. :)

I have another problem which is confuse to me.

Let T be a linear transformation from R^2 into R^2 such that

T(1,1) = (1,0) and T(1,-1) = (0,1)

Find T(1,0) and T(0,2)

okay so I did,

T(1,0) = 1(1,1) + 0(1,-1) = T(1,1) + 0T(1,-1) = (1,0) + 0(0,1) = (0,0)

and

T(0,2) = 0(1,1) + 2(1,-1) = 0T(1,1) + 2T(1,-1) = 0(1,0)+2(0,1) = (0,0)

but the answer for T(1,0) = (1/2, 1/2) and T(1,-1) = (1,-1)

what I did wrong here?

Please help! thanks. - Apr 30th 2007, 04:07 AMtopsquark
The set {(1, 1), (1, -1)} constitutes a basis in R^2 so break each vector up into your basis:

(1, 0) = (1/2)*(1, 1) + (1/2)*(1, -1)

So

T(1, 0) = (1/2)*T(1, 1) + (1/2)*T(1, -1) = (1/2)*(1, 0) + (1/2)*(0, 1) = (1/2, 1/2)

(0, 2) = (1, 1) - (1, -1)

So

T(0, 2) = T(1, 1) - T(1, -1) = (1, 0) - (0, 1) = (1, -1)

-Dan