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Inverse of a special matrix of arbitrary size

  1. Jun 2, 2014 #1
    Hey guys.

    In a project I'm working on, it would be very convienent to express the inverse of this matrix in terms of its size, NxN.

    The matrix is
    [tex]
    \leftbrace \begin{tabular}{c c c c}
    a & b & \ldots & b \\
    b & a & \ldots & b \\
    b & b & \ddots & b \\
    \vdots & vdots & ldots & b \\
    b & b & \ldots & b \\
    \end{tabular}
    \rightbrace
    [/tex]
    [the tex isnt working, but the matrix is just constant b, except on the diagonal where it is a]

    I can see a pattern in the inverses for N=2,3 ; the whole this is divided by det(A) and each element is given by the determinant of its corresponding cominor. This is great because it gives me a recursive formula for computing the inverse. But I'd like to be able to express it explicitly so I can write down the $$i^{th}$$ row in general
     
  2. jcsd
  3. Jun 2, 2014 #2

    D H

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    Staff Emeritus
    Science Advisor

    I think this is what you meant:

    [tex]
    \begin{bmatrix}
    a & b & \ldots & b \\
    b & a & \ldots & b \\
    \vdots & & \ddots & \vdots \\
    b & b & \ldots & a \\
    \end{bmatrix}
    [/tex]

    That's a Toeplitz matrix, and a rather special one at that. For an NxN matrix of this form, the determinant is ##(a-b)^{N-1}(a+(N-1)b)## and the inverse is simply:

    [tex]
    \frac 1{(a-b)(a+(N-1)b)}
    \begin{bmatrix}
    a+(N-2)b & -b & \ldots & -b \\
    -b & a+(N-2)b & \ldots & -b \\
    \vdots & & \ddots & \vdots \\
    -b & -b & \ldots & a+(N-2)b \\
    \end{bmatrix}
    [/tex]
     
  4. Jun 2, 2014 #3

    jbunniii

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    Science Advisor
    Homework Helper
    Gold Member

    Your matrix is of the form
    $$aI + b(1-I)$$
    where ##I## is the ##n \times n## identity matrix, and ##1## is the ##n \times n## matrix consisting of all ones. We might speculate that the inverse has the same form ##cI + d(1-I)##. Let's see if that will work:
    $$\begin{align}
    (aI + b(1-I))(cI + d(1-I)) &= acI + ad(1-I) + bc(1-I) + bd(1-I)^2 \\
    &= (ac - ad - bc)I + (ad + bc)1 + bd(n1 - 1 - 1 + I) \\
    &= (ac - ad - bc)I + (ad + bc)1 + bd((n-2)1 + I) \\
    &= (ac - ad - bc + bd)I + (ad + bc + bd(n-2))1 \\
    \end{align}$$
    where on line 2, we have used the fact that ##1^2 = n1##. We need the final expression to equal the identity ##I##. This will be true provided that ##ac - ad - bc + bd = 1## and ##ad + bc + bd(n-2) = 0##. We have two linear equations with two unknowns (##c## and ##d##), so for most values of ##a## and ##b## there should be a unique solution. I'm too lazy to carry out the rest of the algebra to confirm. :tongue:
     
  5. Jun 2, 2014 #4

    AlephZero

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