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.
Best Answer
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.