Vrftsjtryk

Let $bf A$ be an $N times N$ tri-diagonal matrix $mathbf{A}=frac{1}{{N + 1}}left( {begin{array}{*{20}{c}}
{ - 2}&1&{}&{}\
1&{ - 2}& ddots &{}\
{}& ddots & ddots &1\
{}&{}&1&{ - 2}
end{array}} right)$. How do we find a analytical form of row sum of its inverse $mathbf{A}^{-1}$. Let $s(i)$ denote the sum of elements of the $i$th row in $mathbf{A}^{-1}$, then $s(i)=sum_{j=1}^Nmathbf{A}^{-1}(i,j)$.

Let $s = (s_i)$ be the $Ntimes 1$ matrix of row sums. i.e. the entry at $i^{th}$ row is $s_i = s(i)$.
Let $u$ be the $Ntimes 1$ matrix with all entries $1$. By definition of the row sums, we have
$$s = A^{-1} u quadiffquad As = u$$
Express everyting in terms of entries of $A$ and $s$, this is equivalent to following set of equations:
$$begin{array}{rrrrrrl}
- 2 s_1& + s_2&&&& = N+1\
s_1 & -2s_2& + s_3 &&& = N+1\
&& ddots & ddots & +s_{N} &= N+1\
&&&s_{N-1}&{ -2s_{N}} &= N+1
end{array}tag{*1}$$
Introduce two dummy variables $s_0 = s_{N+1} = 0$. This can be summarized as

$$s_{k-1} - 2s_{k} + s_{k+1} = N+1quadtext{ for } k = 1,ldots, N$$

Notice what's on the left has the form of second order finite difference. To get a constant
on the right. $s_k$ need to be a quadratic polynomial in $k$.
Since $s_0 = s_{N+1} = 0$, we have $s_{k} = alpha k (k - N - 1)$ for some constant $alpha$. One can fix $a$ using any equation in $(*1)$. At the end, we find $alpha = frac{N+1}{2}$ and

$$s(k) = s_k = frac{N+1}{2}k(k - N - 1)$$

The elements of the inverse are

$$
[A^{-1}]_{ij} = (N + 1) begin{cases}
(-1)^{i + j} theta_{i - 1}phi_{j + 1}/theta_N & i < j \
theta_{i - 1}phi_{j + 1}/theta_N & i = j tag{1} \
(-1)^{i + j} theta_{j - 1}phi_{i + 1}/theta_N & i > j \
end{cases}
$$

where $theta$ satisfy the recurrence relation

$$
theta_i = -2theta_{i - 1} - theta_{i - 2}
$$

with initial condition $theta_0 = 1$, $theta_1 = -2$ this leads to

$$
theta_i = (-1)^{i}(i + 1) tag{2}
$$

Similarly

$$
phi_i = -2phi_{i + 1} - phi_{i + 2}
$$

with conditions $phi_{N + 1} = 1$, $phi_N = -2$:

$$
phi_{i} = (-1)^{1 + i - N}(2 - i + N) tag{3}
$$

Replacing in $(1)$, you get

begin{eqnarray}
[A^{-1}]_{ij} &=& (N + 1) begin{cases}
(-1)^{i + j} (-1)^{i - 1}i(-1)^{2 + j - N}(1 - j + N)/(-1)^N(N + 1) & i < j \
(-1)^{i - 1}i(-1)^{2 + j - N}(1 - j + N)/(-1)^N(N + 1) & i = j \
(-1)^{i + j} (-1)^{j - 1}j(-1)^{2 + i - N}(1 - i + N)/(-1)^N(N + 1) & i > j \
end{cases} \ &=&
-begin{cases}
i(1 - j + N) & i < j \
i(1 - j + N) & i = j tag{4}\
j(1 - i + N) & i > j \
end{cases}
end{eqnarray}

Now it is a matter of adding these components

$$bbox[5px,border:2px solid blue]
{
s_i = sum_{j = 1}^N [A^{-1}]_{ij} = begin{cases}
N(N - 7)(N + 1)/2 & 1 leq N leq 8 \
4(N^2 - 6N - 7) & text{otherwise}
end{cases}
}
$$