[VA] SRC#003- New Year 2019 Special
01-28-2019, 05:40 PM
Post: #19
 Thomas Klemm Senior Member Posts: 1,685 Joined: Dec 2013
RE: [VA] SRC#003- New Year 2019 Special
(01-28-2019 12:46 PM)Albert Chan Wrote:  Sadly, I still don't understand the geometric intuition ...

$$v=\begin{bmatrix} 1\\ 1 \end{bmatrix}$$

and a matrix

$$M=\begin{bmatrix} 3 & 0\\ 0 & 2 \end{bmatrix}$$

If we calculate $$M^n v$$ we end up with:

$$M^n v = \begin{bmatrix} 3 & 0\\ 0 & 2 \end{bmatrix}^n \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 3^n & 0\\ 0 & 2^n \end{bmatrix} \begin{bmatrix} 1\\ 1 \end{bmatrix} = \begin{bmatrix} 3^n\\ 2^n \end{bmatrix}$$

Thus the ratio of the coordinates is:

$$\frac{y}{x}=\frac{2^n}{3^n}=\left (\frac{2}{3} \right )^n$$

If we want to know when the $$y$$ component can be neglected compared to the $$x$$ component using a 12-digit calculator we see that's when:

$$\frac{y}{x} < 10^{-12}$$

And therefore when:

$$\left (\frac{2}{3} \right )^n < 10^{-12}$$

Solve that for $$n$$ to get:

$$n > \frac{\log_{10} 10^{-12}}{\log_{10}\frac{2}{3}}=\frac{-12}{\log_{10}2-\log_{10}3}\approx 68.1465$$

Indeed:

2 ENTER 3 ÷ 69 yx
7.0746e-13

But of course we could get a bit closer by using 5e-12 instead of 1e-12.
This leads to $$n>64.1771$$ and indeed $$n=65$$ is enough:

2 ENTER 3 ÷ 65 yx
3.5815e-12

Using the eigenvectors as a basis just adds some coordinate transformations.
We don't know the details (or rather I ignored them) but for a general vector that shouldn't matter much.

But of course this calculation changes if we start with something like:

$$v=\begin{bmatrix} 1\\ 10^k \end{bmatrix}$$

for a big positive value of $$k$$.
But even that would just add $$k$$ to the result.

That's why I consider it only a rough estimate.

Does this make more sense?

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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