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Call a matrix anti-Jordan if it has $-1$s on the subdiagonal, values on the diagonal, and zeroes elsewhere; that is, it is written $$\begin{bmatrix} \lambda_1 & 0 & 0 & \dots \\ -1 & \lambda_2 & 0 & \dots\\ 0 & -1 & \lambda_3 & \dots\\ \vdots &&& \ddots \end{bmatrix}$$ Is there a general formula for the inverse of such a matrix?

If it helps, I'm trying to compute the spectral radius of an $N \times N$ matrix $FV^{-1}$ where $$F = \begin{bmatrix}\beta_1 & \beta_2 & \beta_3 & \dots\\0 & 0 & 0 &\dots\\0 &0 & 0 & \dots\\\vdots &&& \ddots\end{bmatrix}$$ $N$ is "very large" and $$V = \begin{bmatrix} 1 + \mu_1 & 0 & 0 & \dots \\ -1 & 1 + \mu_2 & 0 & \dots\\ 0 & -1 & 1 + \mu_3 & \dots\\ \vdots &&& \ddots \end{bmatrix}.$$

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The inverse is the matrix $B = A^{-1}$ with $b_{ij}=0$ for $i < j$ and $b_{ij} = \lambda_j^{-1}\dots\lambda_i^{-1}$ for $i\ge j$.

Your spectral radius will be the absolute value of $$ \sum_{i=1}^N\frac{\beta_i}{(1+\mu_1)\dots(1+\mu_i)}. $$

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$$\begin{bmatrix} \lambda_1 & 0 & 0 & \dots \\ -1 & \lambda_2 & 0 & \dots\\ 0 & -1 & \lambda_3 & \dots\\ \vdots &\vdots &\vdots & \ddots \end{bmatrix}^{-1}=\begin{bmatrix} \lambda_1^{-1} & 0 & 0 & \dots\\ \lambda_1^{-1}\lambda_2^{-1} & \lambda_2^{-1} & 0 & \dots \\ \lambda_1^{-1}\lambda_2^{-1}\lambda^{-1}_3 & \lambda_2^{-1}\lambda_3^{-1} & \lambda_3^{-1} & \dots\\ \vdots & \vdots & \vdots & \ddots \end{bmatrix}$$ I found this by applying row operations the augmented matrix $[A|I]$, where $A$ is your anti-Jordon matrix and $I$ is the $n\times n$ identity, until it was in the form $[I|B]$.

Edit: My answer was originally wrong, and has been fixed to match amsmath's correct answer.

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    $\begingroup$ This is not correct. When I multiply your $A^{-1}$ to $A$, in the entry $(3,1)$ I get $-\lambda_2^{-1}\lambda_3^{-1}$ insetad of $0$. $\endgroup$ – amsmath Jul 18 '18 at 22:14
  • $\begingroup$ @amsmath Yep, I was too hasty. $\endgroup$ – Mike Earnest Jul 18 '18 at 22:20

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