Application of Inner Product Spaces

how to prove.

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

and

c(u x v) = cu x v = u x cv

thank you.

Quote:

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Originally Posted by test2k6

how to prove.

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

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If * is vector addition and ( x ) is inner product if fails.

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

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

This implies that,
(u*v)=(v*w)

Which is not necessarily true.

Quote:

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Originally Posted by test2k6

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

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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.

Quote:

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Originally Posted by test2k6

c(u x v) = cu x v = u x cv

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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) = cu 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>
= cu x v

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

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 cv

NOTE: in this problem "*" is not being used as the dot product because c is a scalar constant. The symbol "*" simply represents scalar multiplication.

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.

Quote:

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Originally Posted by test2k6

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.

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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