Another Linear Algebra question.

[MetalMat]

If A&B are both orthogonal nxn matricies (basically a square) prove that AB is orthogonal using |ABx| = |x| for all x e R^n

All x's have the arrow sign above them (vectors), and R is the double back R.

 

[Goosemaster]

AxB is cross product btw

 

[StormRider]

The first thing to do is to think what is the definition of an orthogonal matrix.

 

[MetalMat]

Originally posted by: StormRider
The first thing to do is to think what is the definition of an orthogonal matrix.

It is where, when all the rows are dotted into each other each value comes out to zero.

 

[StormRider]

In my old linear algebra text, it says an orthogonal matrix is one where its column vectors form an orthonormal set in R^n. In other words, they are unit vectors that span R^n. (I'm thinking out loud -- it's been a long time since I took linear algebra)

 

[StormRider]

So, yes, each column dotted with another column would be 0. And dotted with itself would be 1.

 

[StormRider]

It's been too long since I took this stuff --

 

[MetalMat]

So you are saying that when you decide to AB, it would equal 1 due to the fact they are both orthogonal, thus any constant vector multiplied into that would take dominance?

 

[StormRider]

So, I'm assuming we have already proven in the past that Q is an orthogonal matrix if and only if |Qx| = |x| for all x in R^n?

 

[MetalMat]

I dont remember the proof exactly, but I sort of remember it. I have a horrible teacher, he is easy but never gives us any homework and I dont really follow him in class.

 

[StormRider]

Try forming the inner product of ABx with itself and see what happens.

 

[StormRider]

Remember ABx is now a vector and you can form the dot product of ABx with itself.

 

[MetalMat]

Im still a little confused, I might have to think about it a little harder when I get home, still at work

 

[StormRider]

Remember that the dot product of a vector with itself is just the square of the norm of the vector. In other words, <v, v> = |v|

 

[StormRider]

I think the key to remember is that if you can show that a matrix M satisfies the property that |Mx| = |x| for any vector x in R^n -- then that matrix must be orthognal.

So, if you show that |ABx| = |x| then the matrix AB must be orthogonal.

Now, remember that ABx is now a vector, so you can form the dot product of it with itself. Try it out and something magical will happen! <ABx, ABx> = ?

 

[MetalMat]

|ABx|

 

[StormRider]

<ABx, ABx> = (ABx)^T (ABx)

The ^T stands for transpose.

 

[MetalMat]

Originally posted by: StormRider
<ABx, ABx> = (ABx)^T (ABx)

The ^T stands for transpose.

But isnt (ABx)^T(ABx) the same as |ABx| in this case?

 

[StormRider]

Yes, but you can manipulate it to get it to equal |x|.

Here's a hint: What does (AB)^T equal to?

 

[StormRider]

What does (ABC) ^T equal to?

 

[StormRider]

Here's another hint. If Q is orthognal, what is Q^T Q equal to?

 

[MetalMat]

Originally posted by: StormRider
Here's another hint. If Q is orthognal, what is Q^T Q equal to?

1

 

[StormRider]

Good, that's correct. It's equal to the Identity Matrix. Now, what does (AB) ^T equal to?

 

[MetalMat]

|AB|

 

[StormRider]

So far, you have

|ABx| = (ABx)^T (ABx) = ?