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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.

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    $\begingroup$ 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. $\endgroup$ Commented Apr 23 at 1:51
  • $\begingroup$ @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. $\endgroup$
    – Hui Zhang
    Commented Apr 30 at 13:25

1 Answer 1

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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 $n=m=2$.

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  • $\begingroup$ I do want $v_0=k\times O(\epsilon)$. But how can one prove the matrix is invertible? $\endgroup$
    – Hui Zhang
    Commented Apr 30 at 13:23
  • $\begingroup$ 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. $\endgroup$
    – PTony
    Commented Apr 30 at 14:52

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