# Inverse of a special triangular matrix

Let $$A$$ be an invertible upper triangular matrix with $$A_{i,j}=A_{i+1,j+1}$$ for all $$i,j$$. How can I show that $$A^{-1}$$ has the same property? That, it is (an upper triangular) matrix with $$A^{-1}_{i,j}=A^{-1}_{i+1,j+1}$$ for all $$i,j$$. I have verified this using computations, but is there a simple proof? Thanks.

The matrix $$A$$ in question is an example of an upper triangular Toeplitz matrix. As noted in the other answer, it can be expressed as $$p(J)$$ for some polynomial $$p$$ in the upper triangular nilpotent Jordan block $$J$$. The converse is also true. In fact, a square matrix is an upper triangular Toeplitz matrix if and only if it is a polynomial in $$J$$.
Now consider your $$A$$. By Cayley-Hamilton theorem, the inverse of $$A$$ must be a polynomial in $$A$$. In turn, $$A^{-1}$$ is a polynomial in $$J$$ and hence it is an upper triangular Toeplitz matrix.
Here's a constructive approach, i.e., one that gives you an algorithm for computing the inverse $$A^{-1}$$. Denote the size of $$A$$ by $$n \times n$$, and denote by $$J := \pmatrix{\cdot&1\\&\cdot&\ddots\\&&\ddots&\ddots\\&&&\cdot&1\\&&&&\cdot}$$ the $$n \times n$$ Jordan block with eigenvalue $$0$$. Then, the condition that $$A$$ is upper triangular and satisfies $$A_{i,j} = A_{i + 1, j + 1}$$ is precisely that $$A = \sum_{k=0}^{n - 1} a_k J^k ,$$ for some constants $$a_0, \ldots, a_{n - 1}$$—respectively the entries of the superdiagonals above the main diagonal—and invertibility just asks that $$a_0 \neq 0$$. As usual, we take the convention that $$J^0 = {\mathbf 1}$$. Write another matrix of that form as $$B = \sum_{k = 0}^{n - 1} b_k J^k$$, $$b_0 \neq 0$$. Then, $$AB = \sum_{k=0}^{n - 1} a_k J^k \sum_{l = 0}^{n - 1} b_l J^l = \sum_{k = 0}^{n - 1} \sum_{l = 0}^{n - 1} a_k b_l J^{k + l}.$$ Reindexing by powers of $$J$$, and using the fact that $$J^n = 0$$, we get $$AB = \sum_{m = 0}^{n - 1} \left(\sum_{k = 0}^m a_k b_{m - k}\right) J^m .$$ So, the condition that $$B$$ is the inverse of $$A$$, i.e., that $$AB = {\mathbf 1}$$, is that the inner sum is $$1$$ for $$m = 0$$ and $$0$$ for $$m > 0$$. In terms of the coefficients, \begin{align*} 1 &= a_0 b_0 \\ 0 &= a_0 b_1 + a_1 b_0 \\ 0 &= a_0 b_2 + a_1 b_1 + a_2 b_0 \\ &\vdots \\ 0 &= a_0 b_{n - 1} + a_1 b_{n - 2} + \cdots + a_{n - 2} b_1 + a_{n - 1} b_0 . \end{align*} Evidently we can successively and uniquely solve the $$i$$th equation for $$b_{i - 1}$$, respectively, giving an explicit formula for $$A^{-1}$$ manifestly in the special upper triangular form.
The first few $$b_i$$ are: \begin{align} b_0 &= \frac1{a_0} \\ b_1 &= -\frac{a_1 b_0}{a_0} = -\frac{a_1}{a_0^2} \\ b_2 &= -\frac{a_1 b_1 + a_2 b_0}{a_0} = -\frac{a_0 a_2 - a_1^2}{a_0^3} . \end{align}