Inverse of a special matrix of arbitrary size

[stormyweathers]

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
<br /> \leftbrace \begin{tabular}{c c c c}<br /> a &amp; b &amp; \ldots &amp; b \\<br /> b &amp; a &amp; \ldots &amp; b \\<br /> b &amp; b &amp; \ddots &amp; b \\<br /> \vdots &amp; vdots &amp; ldots &amp; b \\<br /> b &amp; b &amp; \ldots &amp; b \\<br /> \end{tabular} <br /> \rightbrace <br />
[the tex isn't 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

 

[D H]

I think this is what you meant:

<br /> \begin{bmatrix}<br /> a &amp; b &amp; \ldots &amp; b \\<br /> b &amp; a &amp; \ldots &amp; b \\<br /> \vdots &amp; &amp; \ddots &amp; \vdots \\<br /> b &amp; b &amp; \ldots &amp; a \\<br /> \end{bmatrix} <br />

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:

<br /> \frac 1{(a-b)(a+(N-1)b)}<br /> \begin{bmatrix}<br /> a+(N-2)b &amp; -b &amp; \ldots &amp; -b \\<br /> -b &amp; a+(N-2)b &amp; \ldots &amp; -b \\<br /> \vdots &amp; &amp; \ddots &amp; \vdots \\<br /> -b &amp; -b &amp; \ldots &amp; a+(N-2)b \\<br /> \end{bmatrix} <br />

 

[jbunniii]

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.

 

[AlephZero]

That's a Toeplitz matrix, and a rather special one at that.

Or, you can consider it as a rank-one update to a diagonal matrix, and use the http://en.wikipedia.org/wiki/Sherman–Morrison_formula.