The following equation is given
$$
\begin{bmatrix}
a_1 & b_1\\
b_1 & a_2 & b_2 \\
& b_2 & \ddots & \ddots \\
& & \ddots & \ddots & b_{n-1} \\
& & & b_{n-1} & a_n
\end{bmatrix}\begin{bmatrix}v_1\\v_2\\ \vdots \\v_{n-1} \\v_n\end{bmatrix}=\begin{bmatrix}d_1\\d_2\\ \vdots \\d_{n-1} \\d_n\end{bmatrix}
$$
The $a_i$s and $b_i$s are unknown and two pair of vectors $\{v^{(1)},d^{(1)}\}$ and $\{v^{(2)},d^{(2)}\}$ are given. We need to find $a_i$s and $b_i$s such that the equality holds.
For each row $1, … , n-1$, we face a system of linear equations with two equations and two unknowns which is easy to solve. But for the last row, we have two equations and one unknown $a_n$, because $b_{n-1}$ is already obtained in row $n-1$.
How can I handle this issue?
thanks in advance.
Best Answer
You have a system of $2n$ equations over $2n-1$ variables, so it is over-determined.
As you correctly observed, going top to bottom, you can solve everything uniquely until the last row. There, if both equations will turn out to be multiples of each other, i.e. the implied value of $a_n$ will be the same, your system will have the unique solution.
Otherwise, it will have no solution at all, since $a_n$ cannot be both needed values at the same time.