Finds the analytical form of the row sum of the inverse of a tridiagonal matrix.

linear algebra

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 $s(i)=\sum_{j=1}^N\mathbf{A}^{-1}(i,j)$.

Best Answer

Let $s = (s_i)$ be the $N\times 1$ matrix of row sums. i.e. the entry at $i^{th}$ row is $s_i = s(i)$. Let $u$ be the $N\times 1$ matrix with all entries $1$. By definition of the row sums, we have $$s = A^{-1} u \quad\iff\quad As = u$$ 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

$$s_{k-1} - 2s_{k} + s_{k+1} = N+1\quad\text{ for } k = 1,\ldots, N$$

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

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

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