$M =\begin{bmatrix}40&1\\-1&40\end{bmatrix}$

I've found that the eigenvalues and eigenvectors for this given 2x2 matrix is

$\lambda= \left\{\begin{matrix} 40+i\\ 40-i\\ \end{matrix}\right.$

$v_{1}=[-i,1], v_{2}=[i,1]$

write the start vectors


as a linear combination of the eigenvectors from the matrix M, and use this to show that

$$y_{n}=(40+i)^{n+1}+(40-i)^{n+1}$$ for $$n\geq0$$

This is what I did


I'll skip the calculation, but what I did was rref it which yielded me this as answer

$$RREF(L) =\begin{bmatrix}1&0&40+i\\0&1&40-i\end{bmatrix}$$

which gives me that



Q:but i don't know why and how I should get it for $y_{n}$

ps it also mentions that $x_{n+1}=40x_{n}+y_{n}$, $y_{n+1}=40y_{n}-x_{n}$

EDIT: Forgot to mention this: $y_n$ is the amount of mean beetles and $x_n$ is the amount of nice beetles by the time $t=n$ where the time is in weeks

  • 1
    $\begingroup$ You forgot to say what $y_n$ is. $\endgroup$ – Arnaud Mortier Feb 13 '18 at 15:12
  • $\begingroup$ $y_{n} $ is the amount of mean beetles by the time $t=n$ where the time is in weeks and $x_n$ is the amount of nice beetles $\endgroup$ – user8700908 Feb 13 '18 at 15:19
  • $\begingroup$ I mean how to compute it. I guess that it is via a matrix x vector product, but it should be written. $\endgroup$ – Arnaud Mortier Feb 13 '18 at 15:21
  • $\begingroup$ it only says that i'm going to use those values to show that $y_n=(40+i)^{n+1}+(40-i)^{n+1}$ $\endgroup$ – user8700908 Feb 13 '18 at 15:25

Note that$$\begin{bmatrix}x_0\\y_0\end{bmatrix}=\begin{bmatrix}2\\80\end{bmatrix}=(40+i)\begin{bmatrix}-i\\1\end{bmatrix}+(40-i)\begin{bmatrix}i\\1\end{bmatrix}.$$So\begin{align}\begin{bmatrix}x_n\\y_n\end{bmatrix}&=M^n.\begin{bmatrix}x_0\\y_0\end{bmatrix}\\&=(40+i)^{n+1}\begin{bmatrix}-i\\1\end{bmatrix}+(40-i)^{n+1}\begin{bmatrix}i\\1\end{bmatrix}\end{align}and therefore $y_n=(40+i)^{n+1}+(40-i)^{n+1}$.

  • $\begingroup$ Thanks again, Jose! Can you explain the thought process behind this? $\endgroup$ – user8700908 Feb 13 '18 at 15:30
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    $\begingroup$ @user8700908 There is not much to explain. By definition,$$\begin{bmatrix}x_n\\y_n\end{bmatrix}=M^n.\begin{bmatrix}x_0\\y_0\end{bmatrix}$$and so I computed $M^n.\begin{bmatrix}-i\\1\end{bmatrix}$ and $M^n.\begin{bmatrix}i\\1\end{bmatrix}$. $\endgroup$ – José Carlos Santos Feb 13 '18 at 15:33

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