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I've encountered a problem in which I need to compute the determinant of an almost upper triangular matrix of the following form: $$ A = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\ 1 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\ 1 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\ 1 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\ 1 & 0 & 0 & 0 & a_{5,5} & \\ \vdots & & & & & \ddots \\ \\ 1 & 0 & 0 & 0 & \dots & 0 & a_{N,N} \end{pmatrix} $$

All matrix entries below the diagonal are zero, except those in the first column, which are equal to one.

The matrix is infinite, so $N \to \infty$. I wonder whether there are identities that describe the form of the determinant of this matrix. References to relevant articles are appreciated.

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Note that we can write this matrix in the form $A = B + uv^T$, where $$ B = \begin{pmatrix} 1 & a_{1,2} & a_{1,3} & a_{1,4} & a_{1,5} & \dots \\ 0 & a_{2,2} & a_{2,3} & a_{2,4} & a_{2,5} & \dots \\ 0 & 0 & a_{3,3} & a_{3,4} & a_{3,5} & \dots \\ 0 & 0 & 0 & a_{4,4} & a_{4,5} & \dots \\ 0 & 0 & 0 & 0 & a_{5,5} & \\ \vdots & & & & & \ddots \\ \\ 0 & 0 & 0 & 0 & \dots & 0 & a_{N,N} \end{pmatrix}, \quad u = (0,1,\dots,1)^T, \quad v = (1,0,\dots,0)^T. $$ With the matrix determinant lemma, we find that $$ \det(A) = \det(B + uv^T) = (1 + v^TB^{-1}u) \det(B) \\ = (1 + v^TB^{-1}u) \cdot a_{22} a_{33} \cdots a_{NN}. $$ From there, it suffices to find $v^TB^{-1}u$, i.e. the first entry of $B^{-1}u$.

I don't think that there is a nice explicit form for $v^TB^{-1}u$, but the answer can be computed very efficiently because the matrix is upper triangular.

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  • $\begingroup$ thank you. Maybe the article "On an explicit formula for for inverse triangular matrices" by Baliarsingh and Dutta can help with the determination of the explicit form of $B^{-1}$. Though it seems one needs to compute a lot of determinants with their method, which can get messy $\endgroup$
    – Max Muller
    Aug 14, 2020 at 20:55
  • $\begingroup$ If you encounter any other useful articles on this subject, please let me know! $\endgroup$
    – Max Muller
    Aug 14, 2020 at 20:57

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