Question

Consider the growth of a lilac bush. The state of this lilac bush for several years (at year’s end) is shown in the accompanying sketch. Let n(t) be the number of new branches (grown in the year t) and a(t) the number of old branches. In the sketch, the new branches are represented by shorter lines. Each old branch will grow two new branches in the following year. We assume that no branches ever die.

(a) Find the matrix A such that $\left[\begin{array}{c}n\left(t+1\right)\\ a\left(t+1\right)\end{array}\right]\mathbf{=}\mathit{A}\left[\begin{array}{c}n\left(t\right)\\ a\left(t\right)\end{array}\right]$

(b) Verify that $\left[\begin{array}{c}1\\ 1\end{array}\right]$and $\left[\begin{array}{c}2\\ -1\end{array}\right]$ are eigenvectors of A. Find the associated eigenvalues.

(c) Find closed formulas for n(t) and a(t).

Short answer

The solutions are,

(a)$A=\left[\begin{array}{cc}0& 2\\ 1& 1\end{array}\right]$

(b)${\lambda }_1=2,{\lambda }_2=-1$

(c)$$\begin{gathered} n\left(t\right)=\frac{1}{3}·2^t+\frac{2}{3}·{\left(-1\right)}^t, \\ a\left(t\right)=\frac{1}{3}·2^t+\frac{2}{3}·{\left(-1\right)}^t \end{gathered}$$

Step-by-step solution

Solving for (a):

Let,

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

We have,

$A\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]\Rightarrow a=0,c=1$

We also have

$A\left[\begin{array}{c}0\\ 1\end{array}\right]==\left[\begin{array}{c}2\\ 1\end{array}\right]\Rightarrow b=0,d=1$

Therefore,

$A=\left[\begin{array}{cc}0& 2\\ 1& 1\end{array}\right]$

Solving for (b):

We compute,

$A{v}_1=\left[\begin{array}{cc}0& 2\\ 1& 1\end{array}\right]=\left[\begin{array}{c}1\\ 1\end{array}\right]=\left[\begin{array}{c}2\\ 2\end{array}\right]=2\left[\begin{array}{c}1\\ 1\end{array}\right]$

So,

$v_1=\left[\begin{array}{c}1\\ 1\end{array}\right]$

Is also an eigenvector of A with the eigenvector${\lambda }_1=2$and${\lambda }_2=-1$.

Solving for (c):

We have,

$x_0=\left[\begin{array}{c}1\\ 0\end{array}\right]=\frac{1}{3}\left[\begin{array}{c}1\\ 1\end{array}\right]+\frac{1}{3}\left[\begin{array}{c}2\\ -1\end{array}\right]$

Therefore,

$$\begin{gathered} n\left(t\right)=\frac{1}{3}·2^t+\frac{2}{3}·{\left(-1\right)}^t, \\ a\left(t\right)=\frac{1}{3}·2^t+\frac{1}{3}·{\left(-1\right)}^t \end{gathered}$$