I think that it should not be too hard too justify the claim in the eigenvalue question if one does Exp[Exp[Exp[Exp[Exp[Exp[divisors]]]]]] and Log[Log[Log[Log[Log[Log[eigenvalues]]]]]], infinitely many times. The fastest growing exponentiated number should begin to dominate the the eigenvalues. But I don't know enough about determinants to take the factoring in to account.
Before it goes negative? Ok, I see. The sequence however is the numerators of the Dirichlet series that converges to zero. I might not have described the generating function correctly at the oeis page. It is the translation between symbolical Zeta function products and matrix multiplication that causes me trouble.
@BabakS. I also wanted do the same thing but then I thought that my decision would affect lots of users. Maybe it's better to ponder over the matter for a couple of days in order to make the best decision. :-)
@robjohn: No dear rob. Sometimes, I get bored of some actions of some users. In these cases, I want to leave the site because I can't write them the proper things and want them think about the right way. I am sorry for my self. :(
@robjohn: I wish here was a switch in which I could turn it up or off exactly the same in Facebook.
@BabakS. over 1000 answers ... that means your account is a precious one! Well, to be honest with you, people that don't like us we meet every day in all envirionments. I myself get annoyed sometimes, but I try to go on. You may come here and talk about math if you don't like the main anymore. :-)
@Chris'ssis: Thanks. Your words , both of you, are like pulling water on fire. I don't want to affect others accounts. Thanks rob. I feel I am speaking to my father. Thanks.
@BabakS. I agree with Chris's sis. Rather than deleting your account, you can simply choose not to use it until you feel a bit less oppressed by the others, who are really nothing more than faceless typists if they are not nice.
@BabakS. you know, try to think of those that appreciate your work, and I'm one of those! Many of your answers were really helpful to me, and I plan to retake the study of some of them. :-)
@robjohn: Nice suggestion. Thanks Chris's sis. It is very kind of you. I am nothing here in this vast ocean. I am just a child playing alongside it. OK. I obey you.
@anon I am doing some exercises. I have proven that given two nonzero linear forms in $V^*$, then $\{\varphi_1,\varphi_2\}$ is linearly dependent $\iff$ $\operatorname{ker}\varphi_1=\operatorname{ker}\varphi_2$. Now I am being asked to prove that given $\varphi_1,\ldots,\varphi_r$ linear forms and $\varphi$ another one such that $\bigcap \operatorname{ker}\varphi_i\subseteq \operatorname{ker}\varphi$ then $\varphi\in \langle\varphi_1,\ldots,\varphi_r\rangle$. Here all is finite dimensional.
@PeterTamaroff See this answer and the comments math.stackexchange.com/a/458097/38053. I would just like to know if I'm alone to get spiteful downvotes in retaliation when I comment to point out a mistake. From this specific user I mean.
fix an isomorphism $V\cong V^*$, with $\varphi_i\leftrightarrow\psi_i$. Then $$\bigcap \ker\varphi_i\subseteq\ker\varphi\Leftrightarrow\bigcap \psi_i^\perp=\langle\psi_1,\cdots,\psi_r\rangle^\perp\subseteq\langle \psi\rangle^\perp\Leftrightarrow \langle\psi\rangle\subseteq\langle\psi_1,\cdots\rangle\Leftrightarrow\cdots$$
On the other hand if there is some $x$ not in the other kernel, from $\alpha_1\varphi_1+\alpha_2\varphi_2\equiv 0$ evaluting at $x$ one gets say $\alpha_1=0$; and then since the form is nonzero, we get $\alpha_2=0$.
@PeterTamaroff $\psi_i=\rho(\varphi_i)$ and $\psi=\rho(\varphi)$; like I said in the comment you're replaying to, $\psi$'s are the images of the $\phi$'s under the isomorphism $V^*\to V$
@PeterTamaroff yes, $\tau:V\to V^*:v\mapsto \tau_v$, where $\tau_v\in V^*$
fun fact to keep in mind, isomorphisms $V\to V^*$ are in one-to-one correspondence with inner products on $V$
now, with $\tau(v)=v^*\in V^*$, we have $V\supseteq \ker v^*=\{w\in V:0=v^*(w)=\langle v,w\rangle\}=v^\perp$, the complementary subspace of $\langle v\rangle$ relative to the inner product
FYI, this is all that has been proven in the chapter on the Dual Space: $1.$ For each base there is a unique dual base.
$2.$ If $B=\{v_1,\ldots,v_n\}$ is a base and $B^*=\{\varphi_1,\ldots,\varphi_n\}$ then $(v)_B=(\varphi_1(v),\ldots,\varphi_n(v))$ and $(\varphi)_{B^*}=(\varphi(v_1),\ldots,\varphi(v_n))$.
$3.$ To each base $B_1$ in the dual there is a base $B$ with $B^*=B$.
$4.$ $\dim S^\circ+\dim S=\dim V$
$5.$ $(S+T)^\circ =S^\circ \cap T^\circ$ and $(S\cap T)^\circ=S^\circ+T^\circ$
