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Pseudoinverse - change of basis?

  1. Jun 28, 2012 #1
    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?

    Best regards,
     
  2. jcsd
  3. Jun 28, 2012 #2

    HallsofIvy

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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.
     
  4. Jun 28, 2012 #3
    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! :-)

    Best regards,
     
  5. Jun 28, 2012 #4

    HallsofIvy

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    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.
     
  6. Jun 28, 2012 #5
    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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