math
posted by mathstudent .
There is one step in a proof that I don't understand. Could someone please explain?
u = any vector in vector space S
W = finite dimensional subspace of S with orthonormal basis of vectors {v1, v2, ..., vn}
The theorem to prove is: u can be expressed exactly one way as
u = w1 + w2
w1 = orthogonal projection of u onto W
w2 = component of u orthogonal to W
w = arbitray vector in subspace W
w1 = <u,v1>*v1 + <u,v2>*v2 + ... + <u,vn>*vn
w2 = u  w1
w = k1*v1 + k2*v2 + ... + kn*vn
<w2,w> = <u  w1,w> = <u,w>  <w1,w>
<u,w> = <u,k1*v1 + k2*v2 + ... + kn*vn> = k1*<u,v1> + k2*<u,v2> + ... + kn*<u,vn>
(!!! Don't understand where this equation came from >) <w1,w> = <u,v1>*k1 + <u,v2>*k2 + ... + <u,vn>*kn
<u,w> = <w1,w> so <w2, w> = 0
Where, did that one line come from? the textbook makes a reference to this theorem but includes no other explanation:
If S is an orthornormal basis for an ndimensional inner product space and
us = (u1,u2,...,un)
vs = (v1,v2,...,vn)
then
<u,v> = u1v1 + u2v2 + ... + unvn
Can anyone follow this?
You use the fact that the inner product is bilinear. So, you can always linearly expand. E.g.:
<c_1u_1 + c_2u_2, d_1v_1+d_2v_2> =
c_1d_1<u_1,v_1> +
c_1d_2<u_1,v_2> +
c_2d_1<u_2,v_1> +
c_2d_2<u_2,v_2> +
If you take u_i = v_i to be the ith unit vector, then:
<u_1,v_1> = 1
<u_2,v_2> = 1
<u_1,v_2> = 0
<u_2,v_1> = 0
The vector
c_1u_1 + c_2u_2
can be denoted as
(c_1,c_2)
and
d_1v_1+d_2v_2 as
(d_1,d_2)
The inner product is
c_1d_1 + c_2d_2
because the "cross terms" vanish as they are proportional to inner products of orthogonal vectors.
In your problem:
w1 = <u,v1>*v1 + <u,v2>*v2 + ... + <u,vn>*vn
w = k1*v1 + k2*v2 + ... + kn*vn
and if you take the inner product between w1 and w then you get a sum of the inner products of all the combinations where you pick the ith term from w1 and the jth term from w, summed over i and j. But these are zero unless i = j, because <v_i,v_j> = 0 if i and j are not equal. If i and j are the same then it is 1. So, you end up with the sum of
<u,v_j>k_j from j = 1 to n.
bingo, got it! In hindsight, that's pretty simple. But, I just couldn't figure that out yesterday.
thank you so much count iblis!
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