Proving the Gradient of f(x) in Matrix Notation

In summary, the conversation discusses finding the gradient of a function f(x) and using index notation and the product rule to solve it. The conversation also mentions finding all matrices that satisfy an equation involving the LU factorizations of A and B. The conditions for no solutions are also mentioned.
  • #1
oxxiissiixxo
27
0

Homework Statement


f(x)=(1/2)*(x^T)*(A)*(x)-(x^T)*(b)

Show that the gradient of f(x) is (1/2)*[((A^T)+A)*x]-(b)

where x^transpose is transpose of x and A^transpose is transpose of A.

Note: A is real matrix n*n and b is a column matrix n


Homework Equations





The Attempt at a Solution


I need to kinda proof that.
 
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  • #2
Write it in index notation. (1/2)*x_i*A_ij*x_j+x_i*b_i. All indices summed. Now dx_i/dx_k=delta(i,k). You kinda knew that, right? Use the product rule on the A part.
 
  • #3
Dick said:
Write it in index notation. (1/2)*x_i*A_ij*x_j+x_i*b_i. All indices summed. Now dx_i/dx_k=delta(i,k). You kinda knew that, right? Use the product rule on the A part.

All indices summed. Now dx_i/dx_k=delta(i,k).

what do you mean by that?
 
  • #4
Can you write it in index notation? By dx_i/dx_k=delta(i,k) I mean the derivative is 1 if i=k and 0 if i is not equal k.
 
  • #5
I think that I may be able to make the function into index notation form.
If I make the equation to be (1/2)*x_i*A_ij*x_j+x_i*b_i, I then take the derivative respect with x_k? How this will work?
If k=i, after taking the derivative the equation will become (1/2)*1*A_ij*x_j+1*b_i. What I will have to do for the A_ij? I have not learned how to do the product rule of a matrix yet.
Will k=i and j so that I will have to do 2 derivative of i and j? if so how this works?
 
  • #6
You don't have a matrix anymore once you indexed everything out. There are two x's on the first term one on the right and one on the left which is why you are getting A+A^T.
 
  • #7
would you mind to show me the steps to get from f(x)=(1/2)*(x^T)*(A)*(x)-(x^T)*(b) to (1/2)*[((A^T)+A)*x]-(b).

I did a lot of research about this problem these days. I found a relation: D[f(x)^Tg(x)] = g(x)^Tf'(x) + f(x)^Tg'(x). however i am not convinced. I am sorry this is the first time i do this kinda proof. I can't handle it.
 
  • #8
I can't do that until you at least make a try at the problem. I've told you what to do. The gradient of f(x) is the vector (df(x)/dx_1,df(x)/dx_2,...df(x)/dx_n). Write out the expression as a summation over indices and take d/dx_k of it.
 
  • #9
hi dick thankz i think that i got it YEAH! XD

I have another question if i want to find all the matrices for X that satisfies this equation A*X*B^T = C. How should I deal with this problem? should i also make that into index notation?

B^T is the transpose of B.
 
  • #10
Are A or B invertible?
 
  • #11
It doesn't really say. This is the original question:
Find all matrices X that satisfy the equation A*X*B^T = C, in terms of the LU
factorizations of A and B. State the precise conditions under which there are no
solutions.
 
  • #12
I'm really not an expert on LU factorization stuff. I think you should post this in a new thread so other people can take a crack at it.
 
  • #13
Thanks Man anyway. You are my big help.
 

Related to Proving the Gradient of f(x) in Matrix Notation

What is the gradient of a function in matrix notation?

The gradient of a function in matrix notation is a vector that contains the partial derivatives of the function with respect to each variable. It is often represented as a column vector or row vector in matrix form.

Why is it important to prove the gradient of a function in matrix notation?

Proving the gradient of a function in matrix notation allows us to understand the rate of change of the function in different directions. This is useful in optimization problems and for understanding the behavior of a function.

What are the steps involved in proving the gradient of a function in matrix notation?

The steps involved in proving the gradient of a function in matrix notation include finding the partial derivatives of the function, arranging them in a vector, and representing them in matrix form. Then, the matrix is multiplied by a vector containing the variables, and the resulting expression is simplified to obtain the final gradient vector.

Can the gradient of a function in matrix notation be used to find the direction of maximum increase?

Yes, the gradient of a function in matrix notation can be used to find the direction of maximum increase. The direction of the gradient vector gives the direction of maximum increase, and the magnitude of the gradient vector gives the rate of change in that direction.

Are there any limitations to proving the gradient of a function in matrix notation?

One limitation to proving the gradient of a function in matrix notation is that it only applies to differentiable functions. Also, the gradient may not exist at certain points, such as at discontinuities or sharp corners. Additionally, the gradient may not provide enough information to fully understand the behavior of a function in higher dimensions.

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