[VA] SRC#003- New Year 2019 Special

[Thomas Klemm]

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