Linear Algebra - Definition of OrthonormalDate: 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 |
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