Linear Algebra - Definition of Orthonormal

Date: 10/28/97 at 23:18:29
From: Melanie Wright
Subject: Linear Algebra

Show that the transition matrix P from one orthonormal basis to 
another is unitary, that is, P*P = I.

I can see what a transition matrix is, but I don't understand the 
orthonormal basis portion of this.  Please help!

Date: 10/29/97 at 05:13:28
From: Doctor Mitteldorf
Subject: Re: Linear Algebra

Dear Melanie,

This is a theorem that's true of any vector space, but it's easy to
visualize in a Cartesian plane - we can generalize from there.  

The space we start with is all the points in the plane. To name each 
point, we give its x and y coordinates, measured along horizontal and
vertical axes.  So we name the point (x,y).

But suppose that instead we had used a similar set of axes but rotated
through an angle t. The name for our same point in the new system 
would no longer be (x,y) but (x',y') where

          x' = x cos(t) - y sin(t)
          y' = y cos(t) + x sin(t)

So there's a matrix that takes us from the (x,y) system to the (x',y')
system. We call that matrix P, and in this case we can actually write
it down
          P = { cos(t)   -sin(t) }
              { sin(t)    cos(t) }

Notice that P is unitary. This means that the transpose of P is also
its inverse.  Check that P*P = I

          { cos(t)  sin(t) }    { cos(t) -sin(t) }  =  { 1  0 }
          { -sin(t) cos(t) }    { sin(t)  cos(t) }     { 0  1 }

You may think we're done, and for this 2-D space we really are, but 
this demonstration doesn't generalize very well to other vector 
spaces. It's based on the specific form of P, which in this case must 
be a rotation through a single angle t. Why should P*P = I be true in 
general?

In our familiar basis, the orthonormal basis vectors are (1,0) and 
(0,1). "Ortho" in orthonormal means that (1,0) . (0,1) = 0, and 
"normal" means that (1,0) . (1,0) = 1 and (0,1) . (0,1) = 1.  
In other words, the dot product of any two basis vectors is zero if 
you've chosen two different basis vectors, and 1 if you're taking the 
dot product of any basis vector with itself.

In our new coordinate system, the old basis is no longer (1,0) and 
(0,1). Our new names for these same points are (cos(t),-sin(t)) and
(sin(t),cos(t)). Notice that these new names for the old basis vectors
form the transformation matrix P when you stack them on top of each 
other. This statement IS true in general, and comes right from our 
definitions. Think about it long enough to realize why it is always 
true: for example, the top row of the matrix P tells what happens to 
the vector (1,0) under the transformation.

The final step to the proof is to realize that the dot products of 
these basis vectors with each other still have to be 0 or 1 depending 
on whether they're the same. For example, the dot product of the first 
row of P with itself is 1. But this is exactly the upper left corner 
of the matrix P*P. The next element of P*P to the right, in the (1,2) 
position, is the dot product of the first basis vector with the 
second, (expressed in the new coordinate system).  This must be 0.  
etc.

The form of the proof you'll want to write down depends on the 
notation that you're used to. But the idea is going to be: apply 
the definition of "orthonormal". The statement that the matrix P 
goes from one orthonormal basis to another is exactly the statement 
that P*P = I.

-Doctor Mitteldorf,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/   

Date: 10/29/97 at 15:49:38
From: Anonymous
Subject: Re: Linear Algebra

Dear Doctor Mitteldorf,
     
Thank you so much! You are a great teacher. You spell things out in a 
way that's easy and practical. I actually now understand this stuff.  
I wish you were teaching our class.
     
Cheers!
Melanie