I Simplifying a matrix into an equation

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Im looking to simplify a matrix and express it in a more compact equation form if possible
Hi,

Please see the attached image. I have a matrix and would like to split it up into a nice compact equation if possible. Matrix A seems to be a nice pattern that would lend itself to writing in equation form but I’m not sure what to do. Is it possible? Also do you know how I could correctly separate the column vector out? Would I need to add some padding to this? How would you do that? Many thanks.
 

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PeroK

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It's not clear what you are trying to do. The product of a matrix and a vector is another vector; not another matrix.

It is possible to decompose a matrix into a product of two matrices.
 
I think I’ve found a way to calculate the A part. Is my notation correct? What I am trying to say is take the lower triangular matrix of the Identity matrix, so that it results in all ones of the lower triangular part. Then multiply this by the diagonal matrix of vector x. Please see attached.

Any idea how to progress the a part? Is there something like a hadamard product operation but for a matrix and a column vector? Thanks again :)
 

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PeroK

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I still don't see what equality you are looking for.
 
Appologies I think I should have written J instead of I.

From what I read, the symbol for the lower triangular matrix is L. If I want the non-zero elements of L to be 1 can I write L(J_n)? Is that valid notation to describe that situation?

What I am looking to say with the a part is, take each element in the resulting matrix above and multiply each of the elements by the value given in the column vector with the same row.

eg: [1 2 ;3 4] some operation [10 20]^T = [1*10 2*10; 3*20 4*20]

The Hadamard product is kind of like what I want to do however I am looking for matrices of different sizes.

Is there a way to succinctly say, pad to make the matrices the same size, perform Hadamard product and then remove the unwanted elements?
 
I think Iv found the answer. Is it valid to write L(J_n) though? Thanks again
 

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