[VA] SRC#003- New Year 2019 Special

[Thomas Klemm]

(01-28-2019 12:46 PM)Albert Chan Wrote:  Sadly, I still don't understand the geometric intuition ...

Sorry about that.

Let's start with a vector

\(v=\begin{bmatrix}
1\\
1
\end{bmatrix}\)

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 y^x
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 y^x
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