How do I begin this linear algebra problem?

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Homework Help Overview

The discussion revolves around a linear algebra problem involving the representation of a matrix A such that the product Ax results in a specific vector form. The original poster expresses uncertainty about how to approach the problem, particularly in relation to matrix inverses and the nature of the vector involved.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to relate the problem to matrix inverses, suggesting a method involving the inverse of x. Some participants question the validity of discussing the inverse of a vector, noting that it is not a square matrix. Others suggest focusing on representing A directly as an n x n matrix.

Discussion Status

Participants are exploring different interpretations of how to represent the matrix A. Some guidance has been offered regarding the representation of the transformation, but there is no explicit consensus on the approach. The original poster has indicated progress in understanding the problem.

Contextual Notes

There is a noted confusion regarding the dimensions of the vector and matrix involved, as well as the implications of using inverses in this context. The discussion reflects a learning process with varying levels of understanding among participants.

jp22200
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http://i.imgur.com/MIazUji.png
"Describe the matrix A so that Ax = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1]



Ax = b



I feel like after getting the ball rolling I'll actually be able to work on this problem, but for the time being I haven't the slightest idea how to begin it.

Recently, we've been working on inverses, so I assume that you need to take the inverse of x, and then multiply both sides by the inverse of x so that you're left with A = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1](x^-1), but I don't know exactly how I'd even begin that since it's all variables. Hopefully this counts as "an attempt".

Thank you very much in advance.
 
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jp22200 said:
http://i.imgur.com/MIazUji.png
"Describe the matrix A so that Ax = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1]



Ax = b



I feel like after getting the ball rolling I'll actually be able to work on this problem, but for the time being I haven't the slightest idea how to begin it.

Recently, we've been working on inverses, so I assume that you need to take the inverse of x, and then multiply both sides by the inverse of x so that you're left with A = [x1-x2; x2-x3;...;x(n-1) - x(n); x(n) - x1](x^-1), but I don't know exactly how I'd even begin that since it's all variables. Hopefully this counts as "an attempt".

Thank you very much in advance.
It makes no sense to talk about the inverse of the vector x, since it's not a square matrix. In any case, you don't have to find the inverse of A -- just a representation for A.

Write this:
$$ \begin{bmatrix} x_1 - x_2 \\ x_2 - x_3 \\ . \\ . \\ . \\ x_n - x_1\end{bmatrix}$$
as an n X n matrix, and see where that takes you.
 
Mark44 said:
It makes no sense to talk about the inverse of the vector x, since it's not a square matrix. In any case, you don't have to find the inverse of A -- just a representation for A.

Write this:
$$ \begin{bmatrix} x_1 - x_2 \\ x_2 - x_3 \\ . \\ . \\ . \\ x_n - x_1\end{bmatrix}$$
as an n X n matrix, and see where that takes you.

I must be further behind than I thought, how can I write that as an n X n matrix if it is only an n X 1 matrix to begin with?
 
Gosh, I feel completely daft. Thanks for your help Mark, I figured it out.
 
In general, you can find the matrix representing linear transformation T:U-> V, using ordered basis {u1, u2, ..., un} for U and ordered basis {v1, v[/sub]2[/sub], ..., vm} for V by:

Apply T to u1. Write the result as a linear combination of the basis vectors for V. The coefficients will be the first column. Do the same thing for Tu2 to get the second column and so on.

For example, using the standard basis of Rn for both domain and range spaces, T(1, 0, 0, ..., 0)= (1- 0, 0- 0, 0- 0, ..., 0- 1)= (1, 0, 0, ..., -1) so the first column has 1 in the first row, -1 in the last row and 0 in all other rows.
 

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