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Any notation for component-by-component vector multiplication?

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Curl
#1
Apr23-11, 06:09 PM
P: 757
I have a scalar function and a vector function and I need to make a scalar function as so:

k=[kx ky kz]
T=T(x,y,z)

Function I want:

div(k ? gradT) where "?" would be some operator that multiplies each component of k and gradT to make the vector [kx ∂T/dx , ky ∂T/∂y , kz ∂T/∂z]

that way I can apply the divergence operator and get:
∂/∂x (kx ∂T/∂x) + ∂/∂y (ky ∂T/∂y) + ∂/∂z (kz ∂T/∂z)

So is there some way to express this using elementary notation?
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arsenal997
#2
Apr23-11, 10:09 PM
P: 3
Certeinly there is a notation of that operator, in general if you want to have a operator of that kind you have to put:

[tex]k\nabla\cdot(A)[/tex]

Where A is the vector field that you want to operate.

Greetings
AlexChandler
#3
Apr23-11, 10:37 PM
AlexChandler's Avatar
P: 283
Quote Quote by arsenal997 View Post
Certeinly there is a notation of that operator, in general if you want to have a operator of that kind you have to put:

[tex]k\nabla\cdot(A)[/tex]

Where A is the vector field that you want to operate.

Greetings
k is a vector... what is [tex] \vec k \nabla [/tex] ??

AlexChandler
#4
Apr23-11, 10:51 PM
AlexChandler's Avatar
P: 283
Any notation for component-by-component vector multiplication?

I believe it will be

[tex] \vec \nabla \cdot \left( ( \vec \nabla T)^T \cdot I_3 \vec k \right)^T [/tex]

I think this works
AlexChandler
#5
Apr23-11, 10:53 PM
AlexChandler's Avatar
P: 283
Oh yes... the superscript T means transpose. Treat grad T as a 3 X 1 matrix, and k also as a 3 X 1 matrix. I3 is just the identity matrix for R3.
AlexChandler
#6
Apr23-11, 11:18 PM
AlexChandler's Avatar
P: 283
Actually...

[tex] \left( ( \mathbf{ \nabla } T)^T \mathbf{ I_3 } \textbf{ k } \right) \mathbf{ \nabla } [/tex]

This may be a better way to put it. If we interpret nabla as a 3 X 1 matrix and k is also a 3 X 1 matrix
Curl
#7
Apr24-11, 12:34 AM
P: 757
I can't see what you are trying to do. The Identity matrix does nothing, and multiplying a 1x3 matrix by a 1x3 is not defined.
AlexChandler
#8
Apr24-11, 09:23 AM
AlexChandler's Avatar
P: 283
Quote Quote by Curl View Post
I can't see what you are trying to do. The Identity matrix does nothing, and multiplying a 1x3 matrix by a 1x3 is not defined.
when you multiply k by the identity you get a diagonal matrix with the components of k along the diagonals. Then when you multiply the transpose of the gradient of T by this 3 X 3 matrix you get a 1 X 3 matrix inside of the parenthesis. Then the nabla matrix is 3 X 1 so we have a 1 X 3 multiplied by a 3 X 1 which gives the sum of the products of the components.
AlexChandler
#9
Apr24-11, 09:25 AM
AlexChandler's Avatar
P: 283
Quote Quote by AlexChandler View Post
when you multiply k by the identity you get a diagonal matrix with the components of k along the diagonals. Then when you multiply the transpose of the gradient of T by this 3 X 3 matrix you get a 1 X 3 matrix inside of the parenthesis. Then the nabla matrix is 3 X 1 so we have a 1 X 3 multiplied by a 3 X 1 which gives the sum of the products of the components.
Haha I am sorry you are absolutely right. Let me think about this for a moment
AlexChandler
#10
Apr24-11, 09:59 AM
AlexChandler's Avatar
P: 283
If you can find an operation that will transform k into a diagonal matrix, then the above function should work if you replace I3 k with that operation.
Rasalhague
#11
Apr24-11, 11:37 AM
P: 1,402
You could express it with the dyadic product as

[tex]\nabla \cdot \text{diag}(\textbf{k} \otimes \nabla T) = \nabla \cdot \text{diag}([k] [\nabla T]^T),[/tex]

taking "diag" to mean "form a vector whose components are the diagonal entries of this matrix". Here [k] is a 3x1 matrix (a column vector), and the transpose of [del T] a 1x3 matrix (row vector), so that their product is a 3x3 matrix.

Or simply use the summation sign:

[tex]\sum_{i=1}^{n} \partial_i (k_i \partial_i T) = \sum_{i=1}^{n} \frac{\partial }{\partial x_i}\left ( k_i \frac{\partial T}{\partial x_i} \right ).[/tex]
AlexChandler
#12
Apr24-11, 01:04 PM
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P: 283
Quote Quote by Rasalhague View Post
You could express it with the dyadic product as

[tex]\nabla \cdot \text{diag}(\textbf{k} \otimes \nabla T) = \nabla \cdot \text{diag}([k] [\nabla T]^T),[/tex]

taking "diag" to mean "form a vector whose components are the diagonal entries of this matrix".
Nice! This is the kind of thing I was trying to find. What course can you take to cover these kinds of operations? A second semester of linear algebra?
Rasalhague
#13
Apr25-11, 06:46 AM
P: 1,402
I don't know. I'm learning this stuff from books and the internet, and have never formally taken a course in linear algebra, so I don't know where and when such things would normally be taught. There's a section on dyadics in Snider & Davis: Vector Analysis, 6th edition. Gilbert Strang demonstrates the technique of making an n x n matrix out of vectors by multiplying a column vector by its transpose on the right in one of his MIT linear algebra lectures, which are online here

http://ocw.mit.edu/courses/mathemati...ideo-lectures/

Looking at the titles, it's probably Lecture 16: Projection matrices and least squares, or perhaps Lecture 15: Projections onto subspaces. Another place I've seen something similar is in presentations of the relativistic velocity-addition formula, and that did involve the identity matrix. I thought there was an example on the Wikipedia page: http://en.wikipedia.org/wiki/Velocity_addition But it looks like they've replaced it now; or maybe it was a different page where I saw it. I can post details if you're interested.

The projection matrix idea works like this. Suppose we want to project a vector x onto a vectot a, then we can this as a matrix equation:

[tex]\frac{a^Tx}{a^T a} \; a = \frac{aa^T}{a^T a} \; x = Px.[/tex]

(If a is unit length, we don't need to worry about the denominator.)


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