[VA] SRC#003- New Year 2019 Special
01-16-2019, 09:33 AM (This post was last modified: 01-16-2019 09:36 AM by Thomas Klemm.)
Post: #5
 Thomas Klemm Senior Member Posts: 1,685 Joined: Dec 2013
RE: [VA] SRC#003- New Year 2019 Special
The vector

$$e=\begin{bmatrix} u^2\\ u\\ 1 \end{bmatrix}$$

is eigenvector to the matrix

$$M=\begin{bmatrix} a & u^3 & u^3\\ 1 & a & u^3\\ 1 & 1 & a \end{bmatrix}$$

with eigenvalue $$\lambda=u^2+u+a$$ since

$$\begin{bmatrix} a & u^3 & u^3\\ 1 & a & u^3\\ 1 & 1 & a \end{bmatrix} \begin{bmatrix} u^2\\ u\\ 1 \end{bmatrix}= \begin{bmatrix} au^2+u^4+u^3\\ u^2+au+u^3\\ u^2+u+a \end{bmatrix}= (u^2+u+a) \begin{bmatrix} u^2\\ u\\ 1 \end{bmatrix}$$

For the given values $$a=\pi$$ and $$u^3=2019$$ it appears that $$\lambda$$ is the biggest eigenvalue.
Thus the vector $$M^nv$$ will eventually converge to a multiple of the eigenvector $$e$$ for every vector $$v\neq 0$$.
Since we look only at values in the last column of $$M^n$$ we can just as well calculate:

$$M^n\begin{bmatrix} 0\\ 0\\ 1 \end{bmatrix}$$

Thus the ratio

$$\frac{M^n_{1,3}}{M^n_{2,3}}$$

converges to the value

$$\frac{e_1}{e_2}=\frac{u^2}{u}=u$$

For the given value $$u^3=2019$$ this means $$u=\sqrt[3]{2019}\doteq 12.63898232$$.

Here's a program for the HP-48GX:
Code:
«   @ (n M - ratio )   [ 0 0 1 ] ROT   1 SWAP START     OVER SWAP *   NEXT   SWAP DROP   DUP 1 GET   SWAP 2 GET / »

Example:

With the given matrix M in a variable:

200 M

12.6389823194

Or then using what we know from above:

M EGV DROP { 2 1 } GET INV
(12.6389823194,0)

This works since $$e_1$$ of eigenvectors appears to be 1.

Cheers
Thomas
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 Messages In This Thread [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-15-2019, 10:42 PM RE: [VA] SRC#003- New Year 2019 Special - Paul Dale - 01-16-2019, 03:51 AM RE: [VA] SRC#003- New Year 2019 Special - Carsen - 01-16-2019, 03:57 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Okken - 01-16-2019, 04:24 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-16-2019 09:33 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-16-2019, 09:50 AM RE: [VA] SRC#003- New Year 2019 Special - DavidM - 01-17-2019, 06:04 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 05:04 AM RE: [VA] SRC#003- New Year 2019 Special - DavidM - 01-18-2019, 07:15 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 05:44 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-18-2019, 08:28 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-20-2019, 11:33 AM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-20-2019, 06:13 PM RE: [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-20-2019, 07:40 PM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-09-2019, 01:55 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 01-20-2019, 09:28 PM RE: [VA] SRC#003- New Year 2019 Special - Valentin Albillo - 01-27-2019, 09:38 PM RE: [VA] SRC#003- New Year 2019 Special - pier4r - 01-28-2019, 11:56 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 01-28-2019, 12:46 PM RE: [VA] SRC#003- New Year 2019 Special - pier4r - 01-28-2019, 07:34 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 05:40 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 06:17 PM RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-28-2019, 11:43 PM RE: [VA] SRC#003- New Year 2019 Special - Thomas Klemm - 01-28-2019, 11:59 PM RE: [VA] SRC#003- New Year 2019 Special - rprosperi - 01-29-2019, 02:04 AM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019, 06:46 PM RE: [VA] SRC#003- New Year 2019 Special - Albert Chan - 02-08-2019, 09:36 PM

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