Consider a piece of an (infinitely) large system of uniform tri-diagonal linear
equations:
\begin{eqnarray*}
- a.T_0 + T_1 - b.T_2 = 0 &\quad (1)& \\
- a.T_1 + T_2 - b.T_3 = 0 &\quad (2)& \\
- a.T_2 + T_3 - b.T_4 = 0 &\quad (3)& \\
- a.T_3 + T_4 - b.T_5 = 0 &\quad (4)& \\
- a.T_4 + T_5 - b.T_6 = 0 &\quad (5)&
\end{eqnarray*}
And a boundary condition ($RHS\ne0$) somewhere, but not within our reach.
It is possiblle to use equations (2) and (4) for eliminating $T_2$ and $T_4$
from (3) and replace them by $T_1$ and $T_5$. As follows:
\begin{eqnarray*}
T_2 = a.T_1 + b.T_3 &\quad (2)& \\
T_4 = a.T_3 + b.T_5 &\quad (4)&
\end{eqnarray*}
Giving respectively:
\begin{eqnarray*}
- a (a.T_1 + b.T_3) + T_3 - b (a.T_3 + b.T_5) = 0 & \quad \Longrightarrow \quad & \\
- a^2.T_1 + (1 - 2.a.b) T_3 - b^2.T_5 = 0 & \quad \Longrightarrow \quad & \\
- \left(\frac{a^2}{1-2ab}\right)T_1 + T_3
- \left(\frac{b^2}{1-2ab}\right)T_5 = 0 &\quad (3,2,4)&
\end{eqnarray*}
Thus, by employing this elimination procedure, the matrix pattern:
$$
-a \qquad 1 \qquad -b
$$
has been replaced by:
$$
- \left(\frac{a^2}{1-2ab}\right)
\qquad 0 \qquad 1 \qquad 0 \qquad
- \left(\frac{b^2}{1-2ab}\right)
$$
But we will go even further and eliminate the variables $T_1$ and $T_5$ from
equation (3) by employing (1) and (6) together with (2) and (4):
\begin{eqnarray*}
T_1 = a.T_0 + b.T_2 = a.T_0 + b.(a.T_1 + b.T_3) & \quad (1,2)& \\
(1 - a.b).T_1 = a.T_0 + b^2.T_3 & \quad \Longrightarrow \quad & \\
T_1 = \left(\frac{a}{1-ab}\right) T_0
+ \left(\frac{b^2}{1-ab}\right) T_3 &&
\end{eqnarray*}
\begin{eqnarray*}
T_5 = a.T_4 + b.T_6 = a.(a.T_3 + b.T_5) + b.T_6 & \quad (4,5)& \\
(1 - a.b).T_5 = a^2.T_3 + b.T_6 & \quad \Longrightarrow \quad & \\
T_5 = \left(\frac{a^2}{1-ab}\right) T_3
+ \left(\frac{b}{1-ab}\right) T_6 &&
\end{eqnarray*}
Now substitute (1,2) and (4,5) into (3,2,4), then:
\begin{eqnarray*}
- a^3/(1-2ab)/(1-ab) &T_0& \\
+ \left[1 - 2.a^2.b^2/(1-2ab)/(1-ab)\right] &T_3& \\
- b^3/(1-2ab)/(1-ab) &T_6& = 0
\end{eqnarray*}
And the matrix pattern has become:
$$
- a^3.X(a,b) \qquad 0 \qquad 0 \qquad 1 \qquad 0 \qquad 0 \qquad - b^3.X(a,b)
$$
Where:
$$
X(a,b) = \frac{1/(1-2ab)/(1-ab)}{1 - 2.a^2.b^2/(1-2ab)/(1-ab)} =
$$ $$
\frac{1}{(1-2ab)(1-ab) - 2.a^2.b^2} = \frac{1}{1-3ab}
$$
Hence the matrix pattern, with five equations involved, simplified:
$$
- \left( \frac{a^3}{1-3ab} \right)
\qquad 0 \qquad 0 \qquad 1 \qquad 0 \qquad 0 \qquad
- \left( \frac{b^3}{1-3ab} \right)
$$
It is seen that the elimination procedure with 5 equations involved results
in a pattern which is very much alike the one with 3 equations involved:
$$
- \left(\frac{a^2}{1-2ab}\right)
\qquad 0 \qquad 1 \qquad 0 \qquad
- \left(\frac{b^2}{1-2ab}\right)
$$
This would suggest that an elimination procedure with $2n-1$ equations involved
would result in a matrix pattern like:
$$
- a^n/(1-n.a.b)
\quad 0 \quad ... \quad 0 \quad 1 \quad 0 \quad ... \quad 0 \quad
- b^n/(1-n.a.b)
$$
But this suggestion seems to be false. It is already false for $n=1$, as is clear from
$-a\ne -a/(1-ab)$ and $-b\ne -b/(1-ab)$. And it is false for $n=4$, as is clear
from $4=2\times 2$. Because herewith the function $X(a,b)$ becomes, for $n=4$ :
$$
X(a,b) = \frac{1/(1-2ab)/(1-2ab)}
{1 - 2\left[a^2/(1-2ab)\right]\left[b^2/(1-2ab)\right]}
$$ $$
\frac{1}{(1-2ab)^2 - 2.a^2.b^2} = \frac{1}{1-4ab+2a^2b^2} \ne \frac{1}{1-4ab}
$$
Can we conclude that the elimination of $(2n-1)$ equations from the given uniform
three-diagonal linear system of equations results in a pattern
$$
- a^n/(1-n.a.b)
\quad 0 \quad ... \quad 0 \quad 1 \quad 0 \quad ... \quad 0 \quad
- b^n/(1-n.a.b)
$$
but if and only if $(n=2)$ or $(n=3)$ ?
(linear-algebra) (matrix-decomposition) (gaussian-elimination)