Pseudoinverse - change of basis?

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Discussion Overview

The discussion revolves around the concept of the pseudoinverse and its relationship to changes of basis in the context of solving linear systems, particularly when dealing with matrices where the number of rows is less than the number of columns. Participants explore whether the pseudoinverse can be interpreted as a new basis or if it serves a different purpose in the solution process.

Discussion Character

  • Exploratory, Technical explanation, Debate/contested

Main Points Raised

  • One participant questions if the pseudoinverse can be considered a change of basis, suggesting that it represents a new basis when solving the system Ax = b.
  • Another participant clarifies that a matrix itself is not a basis, emphasizing that a basis consists of a collection of vectors and that the pseudoinverse does not change the basis.
  • A later reply acknowledges the lack of linearly independent vectors in the original problem and reiterates the use of the pseudoinverse to find a solution that is as close as possible to the desired outcome.
  • Further, a participant inquires about the implications of using the pseudoinverse to obtain a minimum norm solution and questions the status of remaining free variables, suggesting they may reside in the null-space.

Areas of Agreement / Disagreement

Participants express differing views on whether the pseudoinverse can be classified as a new basis, with some asserting it cannot be while others propose it might serve that role in the context of solving linear systems. The discussion remains unresolved regarding the interpretation of the pseudoinverse in relation to basis concepts.

Contextual Notes

There are limitations in the discussion regarding the definitions of basis and pseudoinverse, as well as the implications of free variables and null-space in the context of the solutions provided by the pseudoinverse.

kviksand81
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Hello,

I was wondering if the pseudoinverse can be considered a change of basis?

If an m x n matrix with m < n and rank m and you wish to solve the system Ax = b, the solution would hold an infinite number of solutions; hence you form the pseudoinverse by A^T(A*A^T)^-1 and solve for x to get the minimum norm solution. And since A was the original basis the pseudoinverse must be a new basis...? Or am I getting it all wrong?

 
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I don't understand what you mean by "since A was the original basis the pseudoinverse must be a new basis". A matrix is NOT a "basis". A basis is a collection of vectors, not a matrix or linear transformation. A linear transformation is represented by a matrix using a specific basis but finding the generalized inverse does not change the basis.
 
Sorry. From the original problem where m < n, there is not a set of linearly independent vectors to yield a exact solution to the system Ax = b. Then a common technique to solve for a vector that comes as close as possible, would be to form the pseudoinverse, right? This pseudoinverse, is that a new basis or what is it?

Hope that it is more clear what I mean now! :-)

 
No, a basis, for a given vector space, is a set of vectors that span the vector space and are independent so what you are describing is not a basis.
 
Ok.

If I then solve the system Ax =b where the matrix A (m x n) with rank m by means of the right-hand pseudo inverse A^T(A*A^T)^-1, what I get is the minimum norm solution to the m independent columns of A? Or...?

What happens to the remaining free variables? Are they in the null-space?
 

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