Another Linear Algebra question.

[StormRider]

Have you ever seen something like (AB)^T = B^T A^T ?

 

[MetalMat]

Originally posted by: StormRider
Have you ever seen something like (AB)^T = B^T A^T ?

yeah

 

[StormRider]

In other words, the transpose of the product is equal to the product of the transpose of the product of the transposes.

 

[StormRider]

So, what does (ABx)^T equal to?

 

[MetalMat]

[(A)^T(B)^T]x

 

[StormRider]

Not quite. Transpose works for vectors too. If M is a matrix and x is a vector then (Mx) ^T = x^T M^T

 

[MetalMat]

So you a refering to the fact that A^T A is equal to one

 

[MetalMat]

Originally posted by: StormRider
Not quite. Transpose works for vectors too. If M is a matrix and x is a vector then (Mx) ^T = x^T M^T

Ok, I did not remember that.

 

[StormRider]

Also, remember that (ABC)^T = C^T B^T A^T

 

[StormRider]

So, what should (ABx)^T equal to?

 

[MetalMat]

So it would turn into this:

(ABx)^T (ABx) = [A^T B^T x ^T] [A B x] = x^T x

right?

 

[MetalMat]

Originally posted by: StormRider
So, what should (ABx)^T equal to?

A^T B ^T x^T

 

[StormRider]

Originally posted by: MetalMat
So you a refering to the fact that A^T A is equal to one

Yes, you are going to use this fact a little later...

 

[StormRider]

Originally posted by: MetalMat

Originally posted by: StormRider
So, what should (ABx)^T equal to?

A^T B ^T x^T

You forgot to reverse them. It's x^T B^T A^T

 

[StormRider]

Originally posted by: MetalMat
So it would turn into this:

(ABx)^T (ABx) = [A^T B^T x ^T] [A B x] = x^T x

right?

Yes, but you have to reverse the things in the first bracket:

[x ^T B^T A^T][A B x] = ?

 

[MetalMat]

Originally posted by: StormRider

Originally posted by: MetalMat
So it would turn into this:

(ABx)^T (ABx) = [A^T B^T x ^T] [A B x] = x^T x

right?

Yes, but you have to reverse the things in the first bracket:

[x ^T B^T A^T][A B x] = ?

Ax^T B B^T xA^T

 

[StormRider]

And then you multiple the inner stuff to get the identity.

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T I B x = x^T B^T B x = x^T I x = x^T x = |x|

 

[StormRider]

Originally posted by: MetalMat

Originally posted by: StormRider

Originally posted by: MetalMat
So it would turn into this:

(ABx)^T (ABx) = [A^T B^T x ^T] [A B x] = x^T x

right?

Yes, but you have to reverse the things in the first bracket:

[x ^T B^T A^T][A B x] = ?

Ax^T B B^T xA^T

Huh?

 

[MetalMat]

Originally posted by: StormRider
And then you multiple the inner stuff to get the identity.

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T I B x = x^T B^T B x = x^T I x = x^T x = |x|

I dont quite understand how you got from x^T B^T A^T] [A B x] to x^T B^T (A^T A) B x

Thanks for being patient :thumbsup:

 

[MetalMat]

Originally posted by: MetalMat

Originally posted by: StormRider
And then you multiple the inner stuff to get the identity.

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T I B x = x^T B^T B x = x^T I x = x^T x = |x|

I dont quite understand how you got from x^T B^T A^T] [A B x] to x^T B^T (A^T A) B x

Thanks for being patient :thumbsup:

Oh wait wait, duh. you have to go from left to right. Mixing up my laws, long day today.

 

[MetalMat]

Originally posted by: MetalMat

Originally posted by: MetalMat

Originally posted by: StormRider
And then you multiple the inner stuff to get the identity.

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T I B x = x^T B^T B x = x^T I x = x^T x = |x|

I dont quite understand how you got from x^T B^T A^T] [A B x] to x^T B^T (A^T A) B x

Thanks for being patient :thumbsup:

Oh wait wait, duh. you have to go from left to right. Mixing up my laws

I understand. A^T A = 1, so you go to x^T B^T B x -> B^T B =1 -> x^T x = |x|

 

[StormRider]

Matrice multiplication is associative.

So, you can group the inner terms to muliply first. Like with numbers. (CBA) (ABC) = CB(AA)BC

So, I grouped the A^T and A to mulitply to get the identity I

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T (I) B x = x^T (B^T B) x = x^T(I) x = x^T x = |x|

 

[StormRider]

Originally posted by: MetalMat

Originally posted by: MetalMat

Originally posted by: MetalMat

Originally posted by: StormRider
And then you multiple the inner stuff to get the identity.

|ABx| = (ABx)^T (ABx) = [x^T B^T A^T] [A B x] = x^T B^T (A^T A) B x = x^T B^T I B x = x^T B^T B x = x^T I x = x^T x = |x|

I dont quite understand how you got from x^T B^T A^T] [A B x] to x^T B^T (A^T A) B x

Thanks for being patient :thumbsup:

Oh wait wait, duh. you have to go from left to right. Mixing up my laws

I understand. A^T A = 1, so you go to x^T B^T B x -> B^T B =1 -> x^T x = |x|

Yep! You got it!

 

[MetalMat]

Thanks bro 🙂

If you are ever down in Lafayette, Louisiana I'll buy you a beer :beer:

 

[parsley007]

i just wanted to say....good job guys, sh*t looks hard :beer: