I want to see if any pattern that could be formulated exist across the rows (or equivalently , columns) of the matrix inverse (in the middle , let's name it $A^{-1}$) so that an analytical formula that relates $o_n$ and $b_n$ (perhaps along with other b's) exists (for any positive integer n )
$$
\begin{pmatrix}
o_1 \\
o_2 \\
: \\
: \\
: \\
o_n
\end{pmatrix} =
\begin{pmatrix}
1 & 0 & 0 & 0 & … & 0 \\
w & 1 & 0 & 0 & \ddots & 0 \\
i & w & 1 & 0 & \ddots & 0 \\
i & i & w & 1 & \ddots & 0 \\
: & \ddots & \ddots & \ddots & \ddots & \ddots \\
i & i & i & … & w & 1
\end{pmatrix}^{-1}
\begin{pmatrix}
b_1 \\
b_2 \\
: \\
: \\
: \\
b_n
\end{pmatrix}
$$
The matrix inverse $A^{-1}$ is a lower triangular matrix with diagonal full of 1 , subdiagonal full of w and others are i .
when $n = 6$ , the matrix inverse becomes
$$ \left( \begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ -w & 1 & 0 & 0 & 0 & 0 \\ w^2-i & -w & 1 & 0 & 0 & 0 \\ -w^3+2 i w-i & w^2-i & -w & 1 & 0 & 0 \\ w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 & 0 \\ -w^5+4 i w^3-3 i w^2-3 i^2 w+2 i w+2 i^2-i & w^4-3 i w^2+2 i w+i^2-i & -w^3+2 i w-i & w^2-i & -w & 1 \end{array} \right) $$
I recommend matrix calculator https://www.dcode.fr/matrix-inverse to quickly find matrix inverse symbolically .
Since in this case it doesn't matter whether we look at rows or columns , by looking at the first column , I could only write down
$$ \left\{ \begin{array}{ccc}
a_1 & = & 1 \\
a_2 & = & a_1 (-w) \\
a_3 & = & a_2 (-w) – i \\
a_4 & = & a_3 (-w) + iw – i\\
a_5 & = & a_4 (-w) + iw – iw^2 – i + i^2\\
a_6 & = & a_5 (-w) + iw – iw^2 +iw^3 – 2i^2w – i + 2i^2\\
\end{array} \right.
$$
Edit : After reading @metamorphy 's answer
Below is my note ,
since the "1"-diagonal of $S^k$ will lie on the $k^{th}$ subdiagonal , we may visualize $A^{-1}$ and write down
$$
o_n = ( \alpha_+\lambda_{+}^0 + \alpha_- \lambda_{-}^{0} ) b_n +
( \alpha_+\lambda_{+}^{1} + \alpha_- \lambda_{-}^{1} ) b_{n-1} + … +
( \alpha_+\lambda_{+}^{n-1} + \alpha_- \lambda_{-}^{n-1} ) b_{1}
$$
$$
= \sum_{k=0}^{n-1} ( \alpha_+\lambda_{+}^{k} + \alpha_- \lambda_{-}^{k} ) b_{n-k}
$$
with
$$ \alpha_+ = \frac{\lambda_{+} – 1 }{\lambda_{+} – \lambda_{-}} ,\qquad \alpha_- = \frac{\lambda_{-} – 1 }{\lambda_{-} – \lambda_{+}}
$$
Best Answer
Let $S$ be the $n\times n$ matrix with $S_{ij}=\begin{cases}1,&i=j+1\\0,&\text{otherwise}\end{cases}$.
The matrix being inverted is $A=I+wS+iS^2(I-S)^{-1}$, hence $A^{-1}=f(S)$ with $$f(z)=\frac{1-z}{1-(1-w)z-(w-i)z^2}.$$
Next, we find the partial fraction decomposition of $f(z)$, say $$f(z)=\frac{\alpha_+}{1-\lambda_+ z}+\frac{\alpha_-}{1-\lambda_- z},\qquad\lambda_\pm=\frac12\left(1-w\pm\sqrt{(1+w)^2-4i}\right)$$ (this is possible if $(1+w)^2\neq4i$; use continuity or symbolic argument otherwise). Then $$A^{-1}=\sum_{k=0}^{n-1}(\alpha_+\lambda_+^k+\alpha_-\lambda_-^k)S^k$$ because $S^k=0$ for $k\geqslant n$; this gives a closed form of the elements on $k$-th diagonals of $A^{-1}$.