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Orthogonal transformations

by daniel_i_l
Tags: orthogonal, transformations
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daniel_i_l
#1
Jun29-07, 06:29 AM
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1. The problem statement, all variables and given/known data
Given two sub-spaces of R^n - W_1 and W_2 where dimW_1 = dimW_2 =/= 0.
Prove that there exists an orthogonal transformation T:R^n -> R^n so that
T(W_1) = T(W_2)


2. Relevant equations



3. The attempt at a solution
If dimW_1 = dimW_2 = m then we can say that {v_1,...,v_m} is a basis of W_1 and {u_1,...,u_m} is a basis of W_2. We can make from these bases of R^n:
{v_1,...,v_m,a_1,...,a_{n-m}} is a basis of R^n and
{u_1,...,u_m,b_1,...,b_{n-m}} is also.
From these we can make orthonormal bases of R^n (via GS) so that:
{v'_1,...,v'_m,a'_1,...,a'_{n-m}} is an orthonormal basis of R^n and
{u'_1,...,u'_m,b'_1,...,b'_{n-m}} is also.
Now we make a transformation where:
T(v'_i) = u'_i
T(a'_i) = b'_i
Now, since the transformation of an orthonormal base gives another orthonormal base it's a ON transformation. And since {v'_1,...,v'_m} is a basis of W_1 and{u'_1,...,u'_m} is a basis of W_2 then T(W_1) = W_2
Is that a complete proof?
Thanks.
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Dick
#2
Jun29-07, 10:22 PM
Sci Advisor
HW Helper
Thanks
P: 25,228
That's pretty much it. Though you may want to point out that you've diligently applied GS in such a way as to ensure that (v'_1..v'_m) and (u'_1...u'_m) are still a basis for W_1 and W_2. I.e. what order to do GS in? You are probably already thinking of the correct order. I'm just saying this because I can't think of anything else to fault and am wondering why you bothered to post this in the first place. It is really easy.
daniel_i_l
#3
Jun30-07, 01:53 PM
PF Gold
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P: 867
Thanks for the comments :)


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