# Matrix involving reciprocal factorials

Let $$m$$ and $$n$$ be two integers and $$m \le n$$. There are a matrix $$A$$ of $$m$$-by-$$m$$ with $$A(i,j) = 1/(2n+2j-2i)!$$ and a vector $$r$$ of $$m$$ entries with $$r(i) = 2/(2n+2i)!$$.

Is there a formula for the inner product of $$r$$ and the first column of the inverse of $$A$$? Actually, I do not even know whether $$A$$ is invertible (a proof of this is helpful).

Or, can it be shown that the inner product aforementioned is bounded from above by a constant strictly less than one?

The background of this question is as follows. There is a linear system $$\begin{pmatrix}1 & \frac{2}{(2n+2)!} & \frac{2}{(2n+4)!} & \ldots & \frac{2}{(2n+2m)!} \\1 & \frac{1}{(2n)!} & \frac{1}{(2n+2)!} & \ldots & \frac{1}{(2n+2m-2)!}\\ 0 & \frac{1}{(2n-2)!} & \frac{1}{(2n)!} & \ldots & \frac{1}{(2n+2m-4)!}\\ 0 & \frac{1}{(2n-4)!} & \frac{1}{(2n-2)!} & \ldots & \frac{1}{(2n+2m-6)!}\\ \vdots & \vdots & \vdots & \vdots & \vdots \\ 0 & \frac{1}{(2n+2-2m)!} & \frac{1}{(2n+4-2m)!} & \ldots & \frac{1}{(2n)!}\end{pmatrix} \begin{pmatrix}\vphantom{\frac{1}{(2n+2)!}}v_0\\ \vphantom{\frac{1}{(2n+2)!}}v_1\\ \vphantom{\frac{1}{(2n+2)!}}v_2\\ \vphantom{\frac{1}{(2n+2)!}}v_3\\ \vdots\\ \vphantom{\frac{1}{(2n+2)!}}v_m\end{pmatrix} = \begin{pmatrix}\vphantom{\frac{1}{(2n+2)!}}O(\epsilon)\\ \vphantom{\frac{1}{(2n+2)!}}O(\epsilon)\\ \vphantom{\frac{1}{(2n+2)!}}O(\epsilon)\\ \vphantom{\frac{1}{(2n+2)!}} O(\epsilon) \\ \vdots \\ \vphantom{\frac{1}{(2n+2)!}} O(\epsilon)\end{pmatrix},$$ where $$\epsilon>0$$, and $$O(\epsilon)$$ is the big $$O$$ notation for any quantity $$Q$$ for which there exists a constant $$C>0$$ such that $$|Q|< C \epsilon$$. So $$O(\epsilon)$$ can be different values at different places.

It can be found the previous matrix $$A$$ is just the the bottom right part of the above coefficient matrix, from the 2nd column to the last and from the 2nd row to the last. The previous vector $$r$$ is just the top right part of the above coefficient matrix, from the 2nd column to the last in the 1st row.

It is wanted to have an estimate of $$v_0$$. Can it be shown that $$v_0=O(\epsilon)$$ independent of $$m$$ and $$n$$? A less but still wanted goal is just to show the above matrix is invertible.

• The inner product asked for is essentially the ratio of the two minors in the first column of the coefficient matrix, but this seems irrelevant for bounding $v_0$ as seen by applying Cramer's rule. Commented Apr 23 at 1:51
• @Ѕᴀᴀᴅ That is a nice observation. But can one show the ratio you mentioned is less than one? Also, I do not even know whether $A$ is invertible or not. Commented Apr 30 at 13:25

It is clear, by the form of the system of equations itself, that if a solution exists for $$\vec{v}$$, than its first coordinate verifies (or can be chosen in such a way that it verifies) $$v_0 = O(\epsilon)$$.

If your question is (replacing the $$O(\epsilon)$$ with $$\epsilon$$ in the linear equations) whether $$v_0 \sim k \epsilon$$ with $$k$$ less than or equal to $$1$$ for any $$n \geqslant m$$, then the answer is negative. If you solve the simpler (non-singular) system for $$m=2$$.

$$\begin{pmatrix}1 & \frac{2}{(2n+2)!} & \frac{2}{(2n+4)!} & \\1 & \frac{1}{(2n)!} & \frac{1}{(2n+2)!} \\ 0 & \frac{1}{(2n-2)!} & \frac{1}{(2n)!} \end{pmatrix} \begin{pmatrix}\vphantom{\frac{1}{(2n+2)!}}v_0\\ \vphantom{\frac{1}{(2n+2)!}}v_1\\ \vphantom{\frac{1}{(2n+2)!}}v_2\\ \end{pmatrix} = \begin{pmatrix}\vphantom{\frac{1}{(2n+2)!}}\epsilon\\ \vphantom{\frac{1}{(2n+2)!}}\epsilon\\ \vphantom{\frac{1}{(2n+2)!}}\epsilon\\ \end{pmatrix}$$

you get $$v_0=\epsilon (1+\alpha)$$ with $$\alpha>0$$ (but tends to zéro rapidly with $$n$$) so that you cannot, in this case, write $$v_0 \sim \epsilon$$.

Renaming the matrix coefficients $$\begin{pmatrix} 1 & 2a & 2b & \\1 & c & a \\ 0 & d & c \end{pmatrix}$$ , the expression for $$\alpha$$ is given by

$$\displaystyle \alpha=\frac{2 \left(b c-a^2\right)}{a (2 c+d)-2 b d-c^2}$$ which is equal to $$\dfrac{13}{6900}$$ in the case $$n=m=2$$.

• I do want $v_0=k\times O(\epsilon)$. But how can one prove the matrix is invertible? Commented Apr 30 at 13:23
• There is no solution for this system only if the vector on the RHS of the linear system, is not in the range of the matrix. But the symbols $O(\epsilon)$ don't say anythig relevant to the "direction" of this vector. Commented Apr 30 at 14:52