Question

Can someone help me with this proof.. If B is an orthonormal transform, y_1=Bx+2 and y_2=Bx_2 where x_i and y_i are vectors...

Answer 1

If B is an orthonormal transform, y_1=Bx+2 and y_2=Bx_2 where x_i and y_i are vectors. Let \[|x|^{2} = x^{T} x = \sum_{ }^{ } x_{i} ^{2} \] show that \[|y_{1} - y_{2} |^{2} = |x_{1}- x_{2} |^{2} \]

Answer 2

I guess that's supposed to be \(y_1=Bx_1\)?

Answer 3

Yeah sorry y_1=Bx_1

Answer 4

Was no vector space specified?

Answer 5

Nope

Answer 6

I see.
I'll rephrase \(Bx_1\) as \(B(x_1)\) so that it looks similar to function notation.
Since B is an orthonormal transform, it is true that
\[|x_1-x_2|^2=\left<x_1-x_2,x_1-x_2\right>=\left<B(x_1-x_2),B(x_1-x_2)\right>\]right?

Answer 7

How did you get that to that conclusion?

Answer 8

having a hard time grasping what orthonormal transform is

Answer 9

The definition of norm for arbitrary vector spaces is:
\[|u|^2=\left<u,u\right>\]
And an orthonormal transform \(B: V \to V\) is a transform such that
\[\left<u,v\right>=\left<T(u),T(v)\right>\]
Is this different from the definition that you are given?

Answer 10

No no that's very similar to what im given

Answer 11

Okay that make sense then

Answer 12

but so you can insert a constant

Answer 13

into the vector just like that?

Answer 14

I don't get why it's B: V-> V. and then <u,v>=<T(u), T(v)>

Answer 15

Oh you mean <B(u), B(v)>?

Answer 16

Yeah. Sorry bout that.

Answer 17

Yeah no worries okay I get that step you wrote above now

Answer 18

so what's next?

Answer 19

Anyway, we know that \(|x_1-x_2|^2=\left<B(x_1-x_2),B(x_1-x_2)\right>\).
Use the fact that B is a linear transformation to rewrite the RHS as \(\left<B(x_1)-B(x_2),B(x_1)-B(x_2)\right>\).
Finally, replace B(x_1) with y_1 and B(x_2) with y_2 then replace the inner product with the square of a norm.

Answer 20

Oh I see got it hat make sense!

Answer 21

but how do you know b is a linear transform

Answer 22

Because B is an orthonormal transform.

Answer 23

lol okay I guess I have to review my terms..

Answer 24

Thanks very much for your help!

Answer 25

oh but so you don't utilize the transpose part they gave us?

Answer 26

Not really, since it wasn't specified that the vectors come from R^n.

Answer 27

But if it was given do you know how'd you use it to prove it?

Answer 28

the reason is this course does a lot of transposing..

Answer 29

If the specified vector space is R^n, and all of x_1, x_2, y_1, y_2 are column vectors, then you can assume that the standard matrix of B is an n-by-n orthogonal matrix. Call this matrix A. Since A is orthogonal, it is true that \(A^T=A^{-1}\).
Thus, \(B(x_1)=Ax_1=y_1\) and \(B(x_2)=Ax_2=y_2\).
This time, you start from \(|y_1-y_2|^2=(y_1-y_2)^T(y_1-y_2)=(y_1^T-y_2^T)(y_1-y_2)\)
Then replace each y with Ax to get \(((Ax_1)^T-(Ax_2)^T)(Ax_1-Ax_2)\).

Answer 30

Simplify the first part of that last expression to get \[(x_1^TA^T-x_2^TA^T)=(x_1^T-x_2^T)A^T=(x_1^T-x_2^T)A^{-1}=(x_1-x_2)^TA^{-1}\]
So that last one is simplified into
\[(x_1-x_2)^TA^{-1}A(x_1-x_2)=(x_1-x_2)^TI(x_1-x_2)=(x_1-x_2)^T(x_1-x_2)\]
And that last part should look familiar.

Answer 31

Okay everything make sense except the last part, how were you able to just multply (x_1-x_2)^T A^-1 with A(x_1-x_2)?

Answer 32

This was the last expression 3 posts above:
\[\color{red}{((Ax_1)^T-(Ax_2)^T)}(Ax_1-Ax_2)\]
I then showed that
\[\color{red}{((Ax_1)^T-(Ax_2)^T)=(x_1-x_2)^TA^{-1}}\]
Thus, in the end, we get
\[\color{red}{(x_1-x_2)^TA^{-1}}A(x_1-x_2)\]

Answer 33

Ohhh I see what you did now!

Answer 34

Thanks so much for your help, you have no idea how much this clarified things

Answer 35

really appreciate it thanks a lot!

Answer 36

sorry last question what property allows you do (y_1 - y_2) ^T to be (y_1^T - y_2 ^ T)?

Answer 37

never mind I guess it's just a property it's in my notes. Okay thanks again

Answer 38

No problem. :)