Generalising theta transformation formula proof from 1-d to m-d

xfootiecrazeesarax

,,

Way out of my comfort zone, but some thoughts:

Do you have conditions on A? (I think you must, as I think A has to be positive definite or the integral won't converge). If you do, can you justify that A has a square root (e.g. if A is diagonalisable)? If so, then if B^2 = A, I think you'll get a lot of what you want by making the transformation

y = Bx

.

Also, on the latex side,[noparse]

{\bf x}

[/noparse] will give you a bold face

{\bf x}

. (IIRC, the braces are necessary to limit the region put into bold). It's a massive pain to do this whereever needed, so TBH I wouldn't bother if it's reasonably clear that x is supposed to be a vector (as is the case here).

Edit: also, did you mean

\frac{x^tAx}{t} \to x^t x

? In particular the t on the bottom? It doesn't look right. You probably also want to be careful to distinguish transpose from the 't' variable. (I would write

x^Tx

not

x^tx

for clarity).

(edited 7 years ago)

Reply 2

OP

Original post by DFranklin

Edit: also, did you mean

\frac{x^tAx}{t} \to x^t x

? In particular the t on the bottom? It doesn't look right. You probably also want to be careful to distinguish transpose from the 't' variable. (I would write

x^Tx

not

x^tx

for clarity).

Apologies that should be

\frac{x^tAx}{b} \to x^t x

Reply 3

OP

Original post by DFranklin

Way out of my comfort zone, but some thoughts:

Do you have conditions on A? (I think you must, as I think A has to be positive definite or the integral won't converge). If you do, can you justify that A has a square root (e.g. if A is diagonalisable)? If so, then if B^2 = A, I think you'll get a lot of what you want by making the transformation

y = Bx

.
.

Okay, thanks.

Apologies yes A is positive definite.

Okay so because then I'd have

x=B^{-1}y

So

x^{T}Ax/b = (B^{-1}y)^{T}B^2(B^{-1}y)/b

= y^{T}(B^{-1})^{T}BBB^{-1}y/b

= y^{T}y/b

if

(B^{-1})^T=B^{-1}

I am not up to scratch with matrix properties, diagonalising etc, do we a condition that

(B^{-1})^T=B^{-1}

is true?

(I shall now go and have a look into why it can be diagnolized and why this then implies it can be square-rooted)

thanks

Reply 4

Original post by xfootiecrazeesarax

Okay, thanks.

Apologies yes A is positive definite.

Okay so because then I'd have

x=B^{-1}y

So

x^{T}Ax/b = (B^{-1}y)^{T}B^2(B^{-1}y)/b

= y^{T}(B^{-1})^{T}BBB^{-1}y/b

= y^{T}y/b

if

(B^{-1})^T=B^{-1}

I am not up to scratch with matrix properties, diagonalising etc, do we a condition that

(B^{-1})^T=B^{-1}

is true?

(I shall now go and have a look into why it can be diagnolized and why this then implies it can be square-rooted)I'm out of practice enough that you should check all of this, but if A is symmetric, then you should be able to diagonalise A in the form

P\Lambda P^T

. If A is positive definite, all the coefficients of lambda are non-negative, and so you can square root them to form the square root of lambda. Then

B = P \sqrt{\Lambda} P^T

will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).

Reply 5

OP

Original post by DFranklin

I'm out of practice enough that you should check all of this, but if A is symmetric, then you should be able to diagonalise A in the form

P\Lambda P^T

. If A is positive definite, all the coefficients of lambda are non-negative, and so you can square root them to form the square root of lambda. Then

B = P \sqrt{\Lambda} P^T

will do exactly what you want. I think, (in particular B and its inverse will turn out to be symmetric).

okay thank you, so

P

the matrix of eigenvectors normalized is orthonormal so it's inverse is equal to it's transpose.

And the symmetry of B gives

B^2 = BB^T

from which it easily be seen that

B^2 = A

However I'm unsure how to show that

P \sqrt{\Lambda} P^{T}

is orthonormal using the fact that

P

is..?

(edited 7 years ago)

Reply 6

Original post by xfootiecrazeesarax

However I'm unsure how to show that

P \sqrt{\Lambda} P^{T}

is orthonormal using the fact that

P

is..?

It isn't. Why do you need it to be?

(It might be instructive for you to consider what happens in some very simple cases).

E.g. in 1 dimension (so everything is just numbers), we have P = P^T = 1, A is just a real number > 0. For concreteness, suppose A is 3.

Then B ends up being sqrt(3). So B isn't orthonormal (it would have to be +/-1).

But you know the 1D case works, so i don't see this can be an issue.

Reply 7

OP

Original post by DFranklin

It isn't. Why do you need it to be?

.

As post 4
I thought we wanted

B^{-1}=B^{T}

to allow the substitution to bring it to the simplified form

y^{T}y/b

.

edit: apologies I meant orthogonal !
So I.e how to show the statement in the bottom of your post 5 that B and it's inverse are symmetric.

(edited 7 years ago)

Reply 8

Original post by xfootiecrazeesarax

As post 4
I thought we wanted

B^{-1}=B^{T}

to allow the substitution to bring it to the simplified form

y^{T}y/b

.

edit: apologies I meant orthogonal !
So I.e how to show the statement in the bottom of your post 5 that B and it's inverse are symmetric.

To show B is symmetric, start from

B = P \sqrt{\Lambda} P^T

and take the transpose. You end up with

B^T = = P \sqrt{\Lambda} P^T

.

You can do something similar with inv(B).

Reply 9

OP

Original post by DFranklin

To show B is symmetric, start from

B = P \sqrt{\Lambda} P^T

and take the transpose. You end up with

B^T = = P \sqrt{\Lambda} P^T

.

You can do something similar with inv(B).

Ahh yes, thanks.

So now I need the substituion rule of what

dx

will go to in terms of

B

and

dy

on making the substituion

y=Bx

I have never came across a subsitution like this with matrices, and am struggling to find anything online?

Do you have a starting point or a link to some good notes to look at or anything?

Many thanks

Reply 10

OP