Assumption: Given a sequence of square matrices $P^k, k=1,2, ..., n$ with $\lim_{k \to \infty} P^k_{ij} = 0$ for $i \neq j$ (but the diagonal element of $P^k$ may not converge). Moreover, the determinant of the sequence is a constant, suppose $\det P_k = 1, k=1,2, ... n$.
Question: Let $U^k = (P^k)^{-1}$. I am wondering if $\lim_{k \to \infty} U^k_{ij} = 0$, for $i \neq j$?
Currently, I think perhaps we can start from the equaiton: $$ U^k_{ij} = \frac{C^k_{ji}}{\det P^k} $$ , where $C^k_{ji}$ is the $(j, i)$ cofactor of $P^k$. But I can not go through further.
Also, it might by easier if we suppose the diagonal elements of $P^k, k=1,2, ..., n$ are uniformly bounded so that $P^k$ has a convergent subsequence.
No, it turns out. For example, if $\displaystyle P^k = \left( \begin{array}{ccc} \frac{1}{k} & \frac{1}{k} & 0 \\ 0 & \frac{1}{k} & 0 \\ 0 & 0 & k^2 \\ \end{array} \right) $, then $\displaystyle U^k = \left( \begin{array}{ccc} k & -k & 0 \\ 0 & k & 0 \\ 0 & 0 & \frac{1}{k^2} \\ \end{array} \right) $.
On the other hand, if we assume in addition that the diagonal elements of the $P^k$ are uniformly bounded, then the answer is yes. The reason is that each $U_{ij}^k = C_{ij}^k$; each such cofactor entries is a determinant of a matrix that has uniformly bounded entries and one row/column all of whose entries tend to $0$ as $k\to\infty$, and such determinants then must tend to $0$. (To see this, write the determinant as an alternating sum of products of entries of the matrix, or "factor out an $\varepsilon$" from the special row/column.)