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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.

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2 Answers 2

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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.

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  • $\begingroup$ This argument is slick! $\endgroup$ Feb 25 at 3:02
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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}

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