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
$$\leftbrace \begin{tabular}{c c c c} a & b & \ldots & b \\ b & a & \ldots & b \\ b & b & \ddots & b \\ \vdots & vdots & ldots & b \\ b & b & \ldots & b \\ \end{tabular} \rightbrace$$
[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:

$$\begin{bmatrix} a & b & \ldots & b \\ b & a & \ldots & b \\ \vdots & & \ddots & \vdots \\ b & b & \ldots & a \\ \end{bmatrix}$$

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:

$$\frac 1{(a-b)(a+(N-1)b)} \begin{bmatrix} a+(N-2)b & -b & \ldots & -b \\ -b & a+(N-2)b & \ldots & -b \\ \vdots & & \ddots & \vdots \\ -b & -b & \ldots & a+(N-2)b \\ \end{bmatrix}$$

 

[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.

 

[blue_raver22]

.

Hi there,

It's great that you have noticed a pattern in the inverses for different values of N. This is a common approach in mathematics and can be very useful in simplifying calculations. In this case, you have correctly observed that the inverse of the matrix can be expressed in terms of its determinant and the determinants of its corresponding cominors.

To express the inverse explicitly, you can use the formula for the inverse of a matrix of size N. This formula involves finding the adjugate matrix (also known as the adjoint matrix) of the original matrix and then dividing it by the determinant of the original matrix. The adjugate matrix can be found by taking the transpose of the matrix of cofactors, which are the determinants of the cominors.

In summary, to express the inverse of your special matrix of size NxN, you can use the formula for the inverse of a matrix of size N and substitute in the appropriate values for the determinant and the adjugate matrix. This will give you an explicit expression for the inverse in terms of the size N. I hope this helps in your project!