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Inverse of a special matrix

  1. Sep 26, 2014 #1

    srz

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    Hi...can anyone please suggest whether the following inverse has a power series expansion
    [tex](I+\delta A)^{-1}[/tex]
    where [tex]\delta [/tex] is a constant and [tex] A = [/tex]
    [tex] \begin{pmatrix} T & T-1 & T-2 &... & 3 & 2 & 1\\ T-1 & T-1 & T-2 & ... & 3 & 2 & 1 \\ .. \\2 & 2 & 2 &... & 2 & 2 & 1 \\ 1 & 1 & 1 & ... & 1 & 1 & 1 \end{pmatrix} [/tex]
    Thanks!
     
  2. jcsd
  3. Sep 26, 2014 #2

    mathman

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    If delta is small enough.
     
  4. Sep 27, 2014 #3

    srz

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    Thanks. Is it true that [tex] det|I+A|=1+trace(A)+det|A| [/tex]? If not then is there any general expression for [tex] det|I+A| [/tex]
     
  5. Sep 27, 2014 #4
    srz, try A = -I in your first formula.

    There does exist a general formula for det(I+A) in terms of traces of powers of A, but it's rather complicated. Determinant - Wikipedia has it.
     
  6. Sep 27, 2014 #5

    srz

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    Thanks lpetrich.
     
  7. Oct 10, 2014 #6
    For the inverse, you've got ##(I+\delta A)^{-1}=I-\delta A+(\delta A)^2-(\delta A)^3+\ldots##. It should converge if ##\vert\delta A\vert<1##.

    Polynomials and rational functions of a single matrix behave very similarly to the single real variable case.
     
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