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networking – Matrix homeserver (Synapse) setup for mautrix-whatsapp

I would like to install a Matrix homesever on a VPS (running Debian 10) for the sole purpose of using the WhatsApp bridge, mautrix-whatsapp. I really only need to be able to communicate with my personal WhatsApp using a Matrix client, so there’s no need to use federation or anything like that (AFAICT). Since this use is a bit different from common homeserver setups, I was hoping someone could elucidate a few things for me.

The VPS in question mostly serves as an OpenVPN server. I would like to have the Matrix homeserver only accessible from clients connected to the VPN. To the OpenVPN clients, the server is accessible by IP at 192.168.xxx.1. Clients connect to the VPN by IP; there is no externally accessible FQDN associated with it. Is this going to be a problem? N.B. the VPN clients typically use the DNS server (Pi-hole) running on the VPS, which does allow them to reach the server by its hostname. Can I just use this hostname as the domain namee for Synapse and mautrix-whatsapp?
Port 443 is NOT available, because I redirect it to 1194 so clients can connect to the VPN on restrictive public networks. Although the documentation says “a reverse proxy should be reachable publicly with a regular certificate (e.g. Let’s Encrypt) on port 443 that goes to the port 8008 of synapse”, I’m assuming things will work out just fine if I use a different port and point the Matrix client (Element) to it. Is that correct?
In fact, since the client connections will already be encrypted by virtue of OpenVPN, could I forgo the reverse proxy entirely? If so, is that sensible, and is there anything I need to know w.r.t. configuration, etc?
In this configuration, do either Synapse or mautrix-whatsapp require me to open ports to the outside world (i.e. outside of the VPN’s 192.168.xxx.0/24)?

real analysis – Prove or Disprove the following question about power series and Metzler matrix?

Prove or Disprove the following statement:

a) Function $N(s) = 1 + sum_{n=1}^{infty} dfrac{a^ns^n}{(2n)!} $ for a parameter $a in mathbb{R},$ then $N(s)$ satisfies
$$N(s) geq 0 ; forall s geq 0.$$

b) In general, for a matrix $A in M_n(mathbb{R})$ then
$$ N(s) = Id + sum_{n=1}^{infty} dfrac{A^ns^n}{(2n)!} geq 0 quad (text{element-wise})$$
equivalent to $A$ is Metzler, i.e, $a_{ij} geq 0 quad forall i neq j.$

lie groups – Why can the inner product be a criteria for the orthogonality of a matrix

I would like to verify if the matrix
$left( begin{array}{rr} a&b\c&d end{array}right)$ is a Lorentz transformation.

Lorentz Transformations are orthogonal together with the inner product $(x,y) = x_0y_0 -x_iy_i$ . If $a =0$ and $d=0$ then the Matrix is invertible and therefor orthogonal. My script now states that because $(Lx,Lx) leq 0$ for $x = left( begin{array}{r} 1\0 end{array}right)$ L is not a Lorentz transformation for the case that $a=0$.

I would like to understand why the inner product of the resulting vector with itself has to be greater $0$ to be orthogonal?

Function isn’t adding the 2 matrix

Having an issue where my function rotMatfromVec won’t add the 2 Matrices at the end. It takes in 2 inputs, R and theta, where R is a unit vector.

rotMatfromVec[R_, Theta_] :=
  Module[{i, j, k},
    i = {1, 0, 0};
   j = {0, 1, 0};
   k = {0, 0, 1};
   
   iR = Dot[i, R] R;
   jR = Dot[j, R] R;
   kR = Dot[k, R] R;
   mR = N[MatrixForm[{iR, jR, kR}]];
   
   iP = i - iR;
   jP = j - jR;
   kP = k - kR;
   mP = N[MatrixForm[{iP, jP, kP}]];
   
   iC = Cross[R, iP];
   jC = Cross[R, jP];
   kC = Cross[R, kP];
   mC = N[MatrixForm[{iC, jC, kC}]];
   
   mCP = Cos[Theta] mP + Sin[Theta] mC;
   M = MatrixForm[mR + mCP];
   Return[M];
   ];
Print["M = ", rotMatfromVec[i, Pi/2]];

This keeps evaluating to this answer and not performing the addition

{{0,0,0},{0,0,1},{0,-1,0}}+{{1,0,0},{0,0,0},{0,0,0}}

Any idea what I’m doing wrong here?

fa.functional analysis – Lower-bounding the eigenvalues of a certain positive-semidefinite kernel matrix, as a function of the norm of the input matrix

Let $phi:(-1,1) to mathbb R$ be a function such that

$phi$ is $mathcal C^infty$ on $(-1,1)$.
$phi$ is continuous at $pm 1$.

For concreteness, and if it helps, In my specific problem I have $phi(t) := t cdot (pi – arccos(t)) + sqrt{1-t^2}$.

Now, given a $k times d$ matrix $U$ with linearly independent rows, consider the $k times k$ positive-semidefinite matrix $C_U=(c_{i,j})$ defined by $c_{i,j} := K_{phi}(u_i,u_j)$, where

$$
K_phi(x,y) := |x||y|phi(frac{x^top y}{|x||x|})
$$

Question. How express the eigenvalues of $C$ in terms of $U$ and $phi$ ?

I’m ultimated interested in lower-bounding $lambda_{min}(C_U)$ in terms of some norm of $U$ (e.g spectral norm or Frobenius norm).

Let $X$ be the $(d-1)$-dimensional unit-sphere in $mathbb R^d$, equipped with its uniform measure $sigma_{d-1}$, and consider the integral operator $T_phi: L^{2}(X) to L^2(X)$ defined by
$$
T_{phi}(f):x mapsto int K_{phi}(x,y)f(y)dsigma_{d-1}(y).
$$
It is easy to see that $T_phi$ is a compact positive-definite operator.

Question. Are the eigenvalues of $C_U$ be expressed as a function of (eigenvalues of) $K_{phi}$ ?

differential equations – Having problem in solving 11×11 symmetric matrix in mathematica

I am solving an 11×11 matrix in Mathematica. I am facing problem when I try to find the determinant or eigenvalues of this matrix. The error generated shows that the matrix is not square therefore it cannot solve it. But when I manually check the matrix by checking the dimension it gives 11×11 matrix.

Another issue that I am facing is that when I ask the logical question from Mathematica to check whether my matrix is symmetric or not? Mathematica always give false. Whereas, again when I manually check the solution by subtracting the matrix from its transpose, I get a 11×11 null matrix.

So far these things are making me very confused, a help from your end will be highly appreciated. I am also attaching the program with this email.

ClearAll("Global`*")
KLC = {{{(47 C55c)/(30 c) + (6 c (b^2 C11c + a^2 C66c) (Pi)^2)/(
      35 a^2 b^2) + (C11t/a^2 + C66t/b^2) ft (Pi)^2}, {-((7 C55c)/(
       30 c)) + (c (b^2 C11c + a^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {-((4 C55c)/(3 c)) + (
      2 c (b^2 C11c + a^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {2/
      35 (-14 C55c + 
        c^2 (C11c/a^2 + C66c/b^2) (Pi)^2)}, {((6 c (C12c + C66c) + 
        35 (C12t + C66t) ft) (Pi)^2)/(35 a b)}, {(
     c (C12c + C66c) (Pi)^2)/(35 a b)}, {(
     2 c (C12c + C66c) (Pi)^2)/(
     15 a b)}, {-((2 c^2 (C12c + C66c) (Pi)^2)/(
      35 a b))}, {((-22 c C13c + 38 c C55c + 47 C55c ft) (Pi))/(
      60 a c) + (3 c (b^2 C11c + a^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^3 b^2)}, {((Pi) (-6 c^2 C11c fb (Pi)^2 + 
        a^2 (-14 c (C13c + C55c) + 49 C55c fb - (
           6 c^2 (C12c + 2 C66c) fb (Pi)^2)/b^2)))/(420 a^3 c)}, {(
     2 (C13c + C55c) (Pi))/(
     5 a)}}, {{-((7 C55c)/(30 c)) + (
      c (b^2 C11c + a^2 C66c) (Pi)^2)/(35 a^2 b^2)}, {(47 C55c)/(
      30 c) + (6 c (b^2 C11c + a^2 C66c) (Pi)^2)/(
      35 a^2 b^2) + (C11b/a^2 + C66b/b^2) fb (Pi)^2}, {-((4 C55c)/(
       3 c)) + (2 c (b^2 C11c + a^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {(
      4 C55c)/5 - (2 c^2 (b^2 C11c + a^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {(c (C12c + C66c) (Pi)^2)/(
     35 a b)}, {((6 c (C12c + C66c) + 35 (C12b + C66b) fb) (Pi)^2)/(
     35 a b)}, {(2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     2 c^2 (C12c + C66c) (Pi)^2)/(35 a b)}, {(
     6 c^2 C11c ft (Pi)^3 + 
      a^2 (Pi) (14 c (C13c + C55c) - 49 C55c ft + (
         6 c^2 (C12c + 2 C66c) ft (Pi)^2)/b^2))/(
     420 a^3 c)}, {((22 c C13c - 38 c C55c - 47 C55c fb) (Pi))/(
      60 a c) - (3 c (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^3 b^2)}, {-((2 (C13c + C55c) (Pi))/(
      5 a))}}, {{-((4 C55c)/(3 c)) + (
      2 c (b^2 C11c + a^2 C66c) (Pi)^2)/(
      15 a^2 b^2)}, {-((4 C55c)/(3 c)) + (
      2 c (b^2 C11c + a^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {(8 C55c)/(
      3 c) + 16/15 c (C11c/a^2 + C66c/b^2) (Pi)^2}, {0}, {(
     2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     16 c (C12c + C66c) (Pi)^2)/(15 a b)}, {0}, {(
     c^2 C11c ft (Pi)^3 + 
      a^2 (Pi) (-10 (c (C13c + C55c) + C55c ft) + (
         c^2 (C12c + 2 C66c) ft (Pi)^2)/b^2))/(
     15 a^3 c)}, {(2 (c (C13c + C55c) + C55c fb) (Pi))/(3 a c) - (
      c (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      15 a^3 b^2)}, {0}}, {{2/
      35 (-14 C55c + c^2 (C11c/a^2 + C66c/b^2) (Pi)^2)}, {(4 C55c)/
      5 - (2 c^2 (b^2 C11c + a^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {0}, {8/
      105 c (21 C55c + 2 c^2 (C11c/a^2 + C66c/b^2) (Pi)^2)}, {(
     2 c^2 (C12c + C66c) (Pi)^2)/(
     35 a b)}, {-((2 c^2 (C12c + C66c) (Pi)^2)/(35 a b))}, {0}, {-((
      16 c^3 (C12c + C66c) (Pi)^2)/(
      105 a b))}, {-((2 (2 c (C13c + C55c) + 3 C55c ft) (Pi))/(
       15 a)) + (c^2 (b^2 C11c + a^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^3 b^2)}, {-((2 (2 c (C13c + C55c) + 3 C55c fb) (Pi))/(
       15 a)) + (c^2 (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^3 b^2)}, {(8 c (C13c + C55c) (Pi))/(
     15 a)}}, {{((6 c (C12c + C66c) + 35 (C12t + C66t) ft) (Pi)^2)/(
     35 a b)}, {(c (C12c + C66c) (Pi)^2)/(35 a b)}, {(
     2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     2 c^2 (C12c + C66c) (Pi)^2)/(
     35 a b)}, {(47 C44c)/(30 c) + (
      6 c (a^2 C22c + b^2 C66c) (Pi)^2)/(
      35 a^2 b^2) + (C22t/b^2 + C66t/a^2) ft (Pi)^2}, {-((7 C44c)/(
       30 c)) + (c (a^2 C22c + b^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {-((4 C44c)/(3 c)) + (
      2 c (a^2 C22c + b^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {(4 C44c)/
      5 - (2 c^2 (a^2 C22c + b^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {((-22 c C23c + 38 c C44c + 47 C44c ft) (Pi))/(
      60 b c) + (3 c (a^2 C22c + b^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^2 b^3)}, {((Pi) (-6 c^2 C22c fb (Pi)^2 + 
        b^2 (-14 c (C23c + C44c) + 49 C44c fb - (
           6 c^2 (C12c + 2 C66c) fb (Pi)^2)/a^2)))/(420 b^3 c)}, {(
     2 (C23c + C44c) (Pi))/(5 b)}}, {{(c (C12c + C66c) (Pi)^2)/(
     35 a b)}, {((6 c (C12c + C66c) + 35 (C12b + C66b) fb) (Pi)^2)/(
     35 a b)}, {(2 c (C12c + C66c) (Pi)^2)/(
     15 a b)}, {-((2 c^2 (C12c + C66c) (Pi)^2)/(
      35 a b))}, {-((7 C44c)/(30 c)) + (
      c (a^2 C22c + b^2 C66c) (Pi)^2)/(35 a^2 b^2)}, {(47 C44c)/(
      30 c) + (6 c (a^2 C22c + b^2 C66c) (Pi)^2)/(
      35 a^2 b^2) + (C22b/b^2 + C66b/a^2) fb (Pi)^2}, {-((4 C44c)/(
       3 c)) + (2 c (a^2 C22c + b^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {2/
      35 (-14 C44c + c^2 (C22c/b^2 + C66c/a^2) (Pi)^2)}, {(
     6 c^2 C22c ft (Pi)^3 + 
      b^2 (Pi) (14 c (C23c + C44c) - 49 C44c ft + (
         6 c^2 (C12c + 2 C66c) ft (Pi)^2)/a^2))/(
     420 b^3 c)}, {((22 c C23c - 38 c C44c - 47 C44c fb) (Pi))/(
      60 b c) - (3 c (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^2 b^3)}, {-((2 (C23c + C44c) (Pi))/(5 b))}}, {{(
     2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     2 c (C12c + C66c) (Pi)^2)/(15 a b)}, {(
     16 c (C12c + C66c) (Pi)^2)/(
     15 a b)}, {0}, {-((4 C44c)/(3 c)) + (
      2 c (a^2 C22c + b^2 C66c) (Pi)^2)/(
      15 a^2 b^2)}, {-((4 C44c)/(3 c)) + (
      2 c (a^2 C22c + b^2 C66c) (Pi)^2)/(15 a^2 b^2)}, {(8 C44c)/(
      3 c) + 16/15 c (C22c/b^2 + C66c/a^2) (Pi)^2}, {0}, {(
     c^2 C22c ft (Pi)^3 + 
      b^2 (Pi) (-10 (c (C23c + C44c) + C44c ft) + (
         c^2 (C12c + 2 C66c) ft (Pi)^2)/a^2))/(
     15 b^3 c)}, {(2 (c (C23c + C44c) + C44c fb) (Pi))/(3 b c) - (
      c (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      15 a^2 b^3)}, {0}}, {{-((2 c^2 (C12c + C66c) (Pi)^2)/(
      35 a b))}, {(2 c^2 (C12c + C66c) (Pi)^2)/(
     35 a b)}, {0}, {-((16 c^3 (C12c + C66c) (Pi)^2)/(105 a b))}, {(
      4 C44c)/5 - (2 c^2 (a^2 C22c + b^2 C66c) (Pi)^2)/(
      35 a^2 b^2)}, {2/
      35 (-14 C44c + c^2 (C22c/b^2 + C66c/a^2) (Pi)^2)}, {0}, {8/
      105 c (21 C44c + 2 c^2 (C22c/b^2 + C66c/a^2) (Pi)^2)}, {(
      2 (2 c (C23c + C44c) + 3 C44c ft) (Pi))/(15 b) - (
      c^2 (a^2 C22c + b^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^2 b^3)}, {(2 (2 c (C23c + C44c) + 3 C44c fb) (Pi))/(
      15 b) - (c^2 (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^2 b^3)}, {-((8 c (C23c + C44c) (Pi))/(
      15 b))}}, {{((-22 c C13c + 38 c C55c + 47 C55c ft) (Pi))/(
      60 a c) + (3 c (b^2 C11c + a^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^3 b^2)}, {(
     6 c^2 C11c ft (Pi)^3 + 
      a^2 (Pi) (14 c (C13c + C55c) - 49 C55c ft + (
         6 c^2 (C12c + 2 C66c) ft (Pi)^2)/b^2))/(420 a^3 c)}, {(
     c^2 C11c ft (Pi)^3 + 
      a^2 (Pi) (-10 (c (C13c + C55c) + C55c ft) + (
         c^2 (C12c + 2 C66c) ft (Pi)^2)/b^2))/(
     15 a^3 c)}, {-((2 (2 c (C13c + C55c) + 3 C55c ft) (Pi))/(
       15 a)) + (c^2 (b^2 C11c + a^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^3 b^2)}, {((-22 c C23c + 38 c C44c + 47 C44c ft) (Pi))/(
      60 b c) + (3 c (a^2 C22c + b^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^2 b^3)}, {(
     6 c^2 C22c ft (Pi)^3 + 
      b^2 (Pi) (14 c (C23c + C44c) - 49 C44c ft + (
         6 c^2 (C12c + 2 C66c) ft (Pi)^2)/a^2))/(420 b^3 c)}, {(
     c^2 C22c ft (Pi)^3 + 
      b^2 (Pi) (-10 (c (C23c + C44c) + C44c ft) + (
         c^2 (C12c + 2 C66c) ft (Pi)^2)/a^2))/(
     15 b^3 c)}, {(2 (2 c (C23c + C44c) + 3 C44c ft) (Pi))/(15 b) - (
      c^2 (a^2 C22c + b^2 (C12c + 2 C66c)) ft (Pi)^3)/(
      35 a^2 b^3)}, {1/(
      840 a^4 b^4 c) (980 a^4 b^4 C33c + 
        7 a^2 b^2 (a^2 (32 c^2 C44c + 47 C44c ft^2 - 
              4 c (11 C23c ft - 19 C44c ft + 30 ktP)) + 
           b^2 (32 c^2 C55c + 47 C55c ft^2 - 
              4 c (11 C13c ft - 19 C55c ft + 30 ktP))) (Pi)^2 + 
        2 c ft^2 (18 c (b^4 C11c + a^4 C22c + 
              2 a^2 b^2 (C12c + 2 C66c)) + 
           35 (b^4 C11t + a^4 C22t + 
              2 a^2 b^2 (C12t + 2 C66t)) ft) (Pi)^4)}, {1/(
      840 a^4 b^4 c) (140 a^4 b^4 C33c - 
        7 a^2 b^2 (a^2 (8 c^2 C44c - 7 C44c fb ft + 
              2 c (C23c + C44c) (fb + ft)) + 
           b^2 (8 c^2 C55c - 7 C55c fb ft + 
              2 c (C13c + C55c) (fb + ft))) (Pi)^2 - 
        6 c^2 (b^4 C11c + a^4 C22c + 
           2 a^2 b^2 (C12c + 2 C66c)) fb ft (Pi)^4)}, {1/
      15 (-((20 C33c)/c) + 2 c (C44c/b^2 + C55c/a^2) (Pi)^2 + 
        3 ((C23c + C44c)/b^2 + (C13c + C55c)/
           a^2) ft (Pi)^2)}}, {{((Pi) (-6 c^2 C11c fb (Pi)^2 + 
        a^2 (-14 c (C13c + C55c) + 49 C55c fb - (
           6 c^2 (C12c + 2 C66c) fb (Pi)^2)/b^2)))/(
     420 a^3 c)}, {((22 c C13c - 38 c C55c - 47 C55c fb) (Pi))/(
      60 a c) - (3 c (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^3 b^2)}, {(2 (c (C13c + C55c) + C55c fb) (Pi))/(3 a c) - (
      c (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      15 a^3 b^2)}, {-((2 (2 c (C13c + C55c) + 3 C55c fb) (Pi))/(
       15 a)) + (c^2 (b^2 C11c + a^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^3 b^2)}, {((Pi) (-6 c^2 C22c fb (Pi)^2 + 
        b^2 (-14 c (C23c + C44c) + 49 C44c fb - (
           6 c^2 (C12c + 2 C66c) fb (Pi)^2)/a^2)))/(
     420 b^3 c)}, {((22 c C23c - 38 c C44c - 47 C44c fb) (Pi))/(
      60 b c) - (3 c (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^2 b^3)}, {(2 (c (C23c + C44c) + C44c fb) (Pi))/(3 b c) - (
      c (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(15 a^2 b^3)}, {(
      2 (2 c (C23c + C44c) + 3 C44c fb) (Pi))/(15 b) - (
      c^2 (a^2 C22c + b^2 (C12c + 2 C66c)) fb (Pi)^3)/(
      35 a^2 b^3)}, {1/(
      840 a^4 b^4 c) (140 a^4 b^4 C33c - 
        7 a^2 b^2 (a^2 (8 c^2 C44c - 7 C44c fb ft + 
              2 c (C23c + C44c) (fb + ft)) + 
           
           b^2 (8 c^2 C55c - 7 C55c fb ft + 
              2 c (C13c + C55c) (fb + ft))) (Pi)^2 - 
        6 c^2 (b^4 C11c + a^4 C22c + 
           2 a^2 b^2 (C12c + 2 C66c)) fb ft (Pi)^4)}, {1/(
      840 a^4 b^4 c) (980 a^4 b^4 C33c + 
        7 a^2 b^2 (a^2 (32 c^2 C44c + 47 C44c fb^2 - 
              4 c (11 C23c fb - 19 C44c fb + 30 ktP)) + 
           b^2 (32 c^2 C55c + 47 C55c fb^2 - 
              4 c (11 C13c fb - 19 C55c fb + 30 ktP))) (Pi)^2 + 
        2 c fb^2 (18 c (b^4 C11c + a^4 C22c + 
              2 a^2 b^2 (C12c + 2 C66c)) + 
           35 (b^4 C11b + a^4 C22b + 
              2 a^2 b^2 (C12b + 2 C66b)) fb) (Pi)^4)}, {1/
      15 (-((20 C33c)/c) + 2 c (C44c/b^2 + C55c/a^2) (Pi)^2 + 
        3 ((C23c + C44c)/b^2 + (C13c + C55c)/a^2) fb (Pi)^2)}}, {{(
     2 (C13c + C55c) (Pi))/(
     5 a)}, {-((2 (C13c + C55c) (Pi))/(5 a))}, {0}, {(
     8 c (C13c + C55c) (Pi))/(15 a)}, {(2 (C23c + C44c) (Pi))/(
     5 b)}, {-((2 (C23c + C44c) (Pi))/(5 b))}, {0}, {-((
      8 c (C23c + C44c) (Pi))/(15 b))}, {1/
      15 (-((20 C33c)/c) + 2 c (C44c/b^2 + C55c/a^2) (Pi)^2 + 
        3 ((C23c + C44c)/b^2 + (C13c + C55c)/a^2) ft (Pi)^2)}, {1/
      15 (-((20 C33c)/c) + 2 c (C44c/b^2 + C55c/a^2) (Pi)^2 + 
        3 ((C23c + C44c)/b^2 + (C13c + C55c)/a^2) fb (Pi)^2)}, {(
      8 C33c)/(3 c) + 16/15 c (C44c/b^2 + C55c/a^2) (Pi)^2}}};

linear algebra – Trying to determine the determinant of an abstract matrix

I’m trying to write about linear homogeneous recurrence relations and I’ve come up on the following matrix :$$A=begin{pmatrix}
c_1 & c_2 & cdots & c_{k-1} & c_k\
1 & 0 & cdots & 0 & 0\
0 & 1 & ddots & 0 & 0\
vdots & vdots & ddots & ddots & vdots\
0 & 0 & cdots & 1 & 0
end{pmatrix},$$ where $c_1,cdots c_k$ is a real number. I need to find the eigenvalues of this matrix. So far I’ve got that $$c_A(r)=begin{vmatrix}
c_1-t & c_2 & cdots & c_{k-1} & c_k\
1 & -t & cdots & 0 & 0\
0 & 1 & ddots & 0 & 0\
vdots & vdots & vdots & ddots & vdots\
0 & 0 & cdots & 1 & -t
end{vmatrix},$$ I need to express it in its polynomial form but I can’t. I’ve done it for $k=3$ and found $c_A=-r^3+c_1r^2+c_2r +c_3.$ I want to show that $c_A(r)= r^k – sum_{i=1}^{k-1}c_{i}r^{k-i}$.
Any help ?

linear algebra – Pricipal minors for a matrix where all off-diagonal entries are negative

Let $n$ be a positive integer and $Ainmathbb{R}^{ntimes n}$ be a matrix that satisfies:

All the leading principal minors are positive;
All the off-diagonal entries are negative.

Prove that all the principal minors of $A$ are positive.

I have a very indirect solution to this problem, and I’m still looking for a direct approach that works.

numerical integration – Matrix as the discretization of a function $u$ over a rectangle in 3D, how to integrate fast along the axes

a minimal example for my problem

(* domain*)
nn = 10.;
{xmin, ymin, zmin} = {0, 0, 0};
{xmax, ymax, zmax} = {1., 1., 1.};
xx = Subdivide(xmin, xmax, nn);
yy = Subdivide(ymin, ymax, nn);
zz = Subdivide(zmin, zmax, nn);
dx = xx((2)) - xx((1));
(* create discrete uu *)
u(x_, y_, z_) := dx*(x + y + z)
uu = Table(u(x, y, z), {x, 0, nn}, {y, 0, nn}, {z, 0, nn})
(* create discrete uu *)
u(x_, y_, z_) := dx*(x + y + z)
uu = Table(u(x, y, z), {x, 0, nn}, {y, 0, nn}, {z, 0, nn})

So I have a function $u$ from 3D to 1D approximated in a list uu and I want to calculate e.g.
$int_0^1 xcdot u(x,y,z)dx$ or $int_0^1int_0^1int_0^1 u(x,y,z)dxdydz$.
How do I do this fast?
My search begun at Listintegrate (1) and then I arrived at (2) which brought me to examples like

data = {1, 2, 3}
f = Interpolation(data((1)), InterpolationOrder -> 1)
Plot(f(x), {x, 1, 3})
NIntegrate(f(x), {x, 1, 3})
Integrate(f(x), {x, 1, 3})

but is this really the way to go?
Thanks in advance for all the tipps

(1) https://reference.wolfram.com/language/FunctionApproximations/ref/ListIntegrate.html
(2) https://reference.wolfram.com/language/ref/Interpolation.html

rotation – Matrix 3d translation turtle logo maths

I’m trying to write a 3d compatible turtle/logo program, so that I can feed a test of commands into a class, and then paint the 3d location that the turtle walks through. Ideally it should handle any vector or direction to move through.

I’m using this matrix class, as I think this is putting me on the right path:

https://github.com/CoolCat467/Gameobjects-Python-3/blob/master/gameobjects/matrix44.py

I’d like to be able to feed a set of comands such as:

Move forwards
Turn Left
Move forwards
Turn right
Move forwards

But when I’m trying things out, I’m getting big long sweeping curves, rather than making the tighter angles and turns I’d like to happen:

for i in range(0, 10):
    identity.up += Matrix44.z_rotation(radians(45)).get_row_vec3(1)
    turtle_heading = Vector3(identity.up)
    identity.translate = identity.translate + turtle_heading * 2
    bpy.ops.mesh.primitive_cube_add(location=identity.get_row_vec3(3), size=1)

How would I get the matrix to peform in a more turtle like way (And have I understood how I should be using the matrix correctly?)

If I change the angle per step, it seems to work..

But then breaks when it goes over 360 degrees

These are the angles if I reset to 0 after 360:

45
90
135
180
225
270
315
0
45

and this if I just keep going

45
90
135
180
225
270
315
360
405

Both produce the same result, but it looks like it’s working up until 360 degrees, but then something breaks apart