What is inverse of a matrix whose diagonal elements are all zero.

Question

I am a science researcher and I got a problem to find the generic inverse of the following matrix:

$$ A_n = \left(\begin{array}{ccc} 0 & a_2 & a_3 & ... & a_{n-1} & a_n \\ a_1 & 0 & a_3 & ... & a_{n-1} & a_n \\ a_1 & a_2 & 0 & ... & a_{n-1} & a_n \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ a_1 & a_2 & a_3 & ... & 0 & a_n \\ a_1 & a_2 & a_3 & ... & a_{n-1} & 0 \\ \end{array}\right) $$

I figured out that for n=2,3

$$ A_2^{-1} = \left(\begin{array}{ccc} 0 & a_2^{-1} \\ a_1^{-1} & 0 \\ \end{array}\right) $$

$$ A_3^{-1} = \frac{1}{2} \left(\begin{array}{ccc} -a_1^{-1} & a_1^{-1} & a_1^{-1} \\ a_2^{-1} & -a_2^{-1} & a_2^{-1} \\ a_3^{-1} & a_3^{-1} & -a_3^{-1} \\ \end{array}\right) $$

but can we extend it to a general case? Can anyone help?? Thanks!

[later]

It looks like it is

$$ m_{ij} = -\frac{n-2}{n-1} a_{i}^{-1}\ (i=j) $$ $$ = \frac{1}{n-1} a_{i}^{-1}(else) $$

Answer 1

Your matrix can be written as $D_n + u_n \, v_n^T$:

$$u_n = \left[ \begin{matrix} 1 & ... & 1 \end{matrix} \right]^T$$

$$v_n = \left[ \begin{matrix} a_1 & ... & a_n \end{matrix} \right]^T$$

$$D_n = - \text{diag}(v_n) = - \left[ \begin{matrix} a_1 & ... & 0 \\ \vdots & ... & \vdots \\ 0 & ... & a_n \\ \end{matrix} \right]$$

Which reminds me of the Sherman–Morrison formula:

$$\left(A + u \, v^T\right)^{-1} = A^{-1} - {A^{-1} \, u \, v^T \, A^{-1} \over 1 + v^T \, A^{-1} \, u}$$

[EDIT]

Your pattern seems to be right. Defining $w_n = \left[ \begin{matrix} a_1^{-1} & ... & a_n^{-1} \end{matrix} \right]^T$:

$$ $$

$$1 + v^T \, A^{-1} \, u = 1-n$$

$$A^{-1} \, u = - w_n$$

$$ v^T \, A^{-1} = - u_n^T$$

$$\left(A + u \, v^T\right)^{-1} = - \text{diag}(w_n) - { w_n \, u_n^T \over 1-n}$$

$$ \left(A + u \, v^T\right)^{-1} = - \left[ \begin{matrix} a_1^{-1} & ... & 0 \\ \vdots & ... & \vdots \\ 0 & ... & a_n^{-1} \\ \end{matrix} \right] - \frac{1}{1-n} \left[ \begin{matrix} a_1^{-1} & ... & a_1^{-1} \\ \vdots & ... & \vdots \\ a_n^{-1} & ... & a_n^{-1} \\ \end{matrix} \right] $$

$$ $$

$$ m_{ij} = -a_{i}^{-1} -\frac{a_{i}^{-1}}{1-n} = \frac{n-2}{1-n} \, a_{i}^{-1} = - \frac{n-2}{n-1} \, a_{i}^{-1} \, (i=j) $$ $$ = -\frac{a_{i}^{-1}}{1-n} = \frac{1}{n-1} \, a_{i}^{-1} \, (else) \hphantom{aaaaaaaaaaaaaaa} $$

Answer 2

Your pattern extends to any $n$. You can see this by multiplying out:

$$\begin{pmatrix} -(n-2)a_1^{-1} & a_1^{-1} & ... & a_1^{-1} \\ a_2^{-1} & -(n-2)a_2^{-1} & ... & a_2^{-1} \\ ... & ... & ... & ... \\ ... & ... & ... & ... \\ a_n^{-1} & a_n^{-1} & ... & -(n-2)a_n^{-1} \end{pmatrix} \begin{pmatrix} 0 & a_2 & a_3 & ... & a_{n-1} & a_n \\ a_1 & 0 & a_3 & ... & a_{n-1} & a_n \\ ... & ... & ... & ... & ... & ... \\ ... & ... & ... & ... & ... & ... \\ a_1 & a_2 & a_3 & ... & a_{n-1} & 0 \end{pmatrix}=(n-1)I$$

For $i=j$, there are $(n-1)$ $a_i^{-1}$ in the $i$th row each multiplied by $a_i$ in the $j$th ($=i$) column to give $n-1$.

For $i\ne j$, one of $a_i^{-1}$ is multiplied by $0$; so in fact there are $(n-2)a_i^{-1}a_j$, canceled out by a single $(n-2)a_i^{-1}a_j$.