Diagonalisation and powers Problem set A One question at a time

Write each matrix in the form \(PDP^{-1}\).

1. (a) \(\begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\)

   • \(\lambda = 5 \text{ with } \mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix};\quad \lambda = 2 \text{ with } \mathbf{v} = \begin{pmatrix}2 \\ -1\end{pmatrix}\)
   • \(P = \begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix},\quad D = \begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix}\)
   • \(\text{Check: } M = PDP^{-1} = \begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix}\begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix}\begin{pmatrix}\frac{1}{3} & \frac{2}{3} \\ \frac{1}{3} & -\frac{1}{3}\end{pmatrix} = \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\)

2. (b) \(\begin{pmatrix}1 & 2 \\ 4 & 3\end{pmatrix}\)

   • \(\lambda = 5 \text{ with } \mathbf{v} = \begin{pmatrix}1 \\ 2\end{pmatrix};\quad \lambda = -1 \text{ with } \mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}\)
   • \(P = \begin{pmatrix}1 & 1 \\ 2 & -1\end{pmatrix},\quad D = \begin{pmatrix}5 & 0 \\ 0 & -1\end{pmatrix}\)
   • \(\text{Check: } M = PDP^{-1} = \begin{pmatrix}1 & 1 \\ 2 & -1\end{pmatrix}\begin{pmatrix}5 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix}\frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & -\frac{1}{3}\end{pmatrix} = \begin{pmatrix}1 & 2 \\ 4 & 3\end{pmatrix}\)

For \(M = \begin{pmatrix}3 & 2 \\ 1 & 4\end{pmatrix}\), find \(M^n\) in terms of \(n\), and hence find \(M^4\).

• \(\text{From question 5.6: } \lambda = 5, \mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix};\; \lambda = 2, \mathbf{v} = \begin{pmatrix}2 \\ -1\end{pmatrix}.\)
• \(P = \begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix},\; P^{-1} = \frac{1}{-3}\begin{pmatrix}-1 & -2 \\ -1 & 1\end{pmatrix}\)
• \(M^n = PD^nP^{-1} = \begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix}\begin{pmatrix}5^n & 0 \\ 0 & 2^n\end{pmatrix}\cdot\tfrac{1}{3}\begin{pmatrix}1 & 2 \\ 1 & -1\end{pmatrix}\)
• \(M^n = \frac{1}{3}\begin{pmatrix}5^n + 2(2^n) & 2(5^n) - 2(2^n) \\ 5^n - 2^n & 2(5^n) + 2^n\end{pmatrix}\)
• \(n = 4: \; 5^4 = 625,\; 2^4 = 16.\)
• \(M^4 = \frac{1}{3}\begin{pmatrix}625 + 32 & 1250 - 32 \\ 625 - 16 & 1250 + 16\end{pmatrix} = \begin{pmatrix}219 & 406 \\ 203 & 422\end{pmatrix}\)

\(M = \begin{pmatrix}2 & 3 \\ 4 & 1\end{pmatrix}\) has eigenvalues 5 and −2. Diagonalise \(M\) and use this to find \(M^5\).

• \(\lambda = 5: \; -3x + 3y = 0 \text{ gives } \mathbf{v} = \begin{pmatrix}1 \\ 1\end{pmatrix}.\)
• \(\lambda = -2: \; 4x + 3y = 0 \text{ gives } \mathbf{v} = \begin{pmatrix}3 \\ -4\end{pmatrix}.\)
• \(P = \begin{pmatrix}1 & 3 \\ 1 & -4\end{pmatrix},\; D = \begin{pmatrix}5 & 0 \\ 0 & -2\end{pmatrix},\; P^{-1} = \frac{1}{-7}\begin{pmatrix}-4 & -3 \\ -1 & 1\end{pmatrix}\)
• \(M^5 = PD^5P^{-1} \text{ with } D^5 = \begin{pmatrix}3125 & 0 \\ 0 & -32\end{pmatrix}.\)
• \(M^5 = \begin{pmatrix}1 & 3 \\ 1 & -4\end{pmatrix}\begin{pmatrix}3125 & 0 \\ 0 & -32\end{pmatrix}\begin{pmatrix}\frac{4}{7} & \frac{3}{7} \\ \frac{1}{7} & -\frac{1}{7}\end{pmatrix} = \begin{pmatrix}1772 & 1353 \\ 1804 & 1321\end{pmatrix}\)

A matrix \(M\) has eigenvalue \(4\) with eigenvector \(\begin{pmatrix}1 \\ 2\end{pmatrix}\) and eigenvalue \(1\) with eigenvector \(\begin{pmatrix}1 \\ 1\end{pmatrix}\).

1. (a) Find \(M\).

   • \(P = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix},\; D = \begin{pmatrix}4 & 0 \\ 0 & 1\end{pmatrix},\; P^{-1} = \begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}\)
   • \(M = PDP^{-1} = \begin{pmatrix}4 & 1 \\ 8 & 1\end{pmatrix}\begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix} = \begin{pmatrix}-2 & 3 \\ -6 & 7\end{pmatrix}\)

2. (b) Find \(M^{3}\).

   • \(M^{3} = PD^{3}P^{-1} = \begin{pmatrix}1 & 1 \\ 2 & 1\end{pmatrix}\begin{pmatrix}64 & 0 \\ 0 & 1\end{pmatrix}\begin{pmatrix}-1 & 1 \\ 2 & -1\end{pmatrix}\)
   • \(= \begin{pmatrix}-62 & 63 \\ -126 & 127\end{pmatrix}\)

Each year 20% of the people living in a town move to the countryside around it, and 30% of the people in the countryside move into the town. This is modelled by \(T = \begin{pmatrix}0.8 & 0.3 \\ 0.2 & 0.7\end{pmatrix}\). At the start there are 5000 people in each. (a) Find the numbers after 3 years. (b) Find the matrix that \(T^n\) approaches, and the numbers in the long term.

• \(\text{From question 5.6: } \lambda = 1, \mathbf{v} = \begin{pmatrix}3 \\ 2\end{pmatrix};\; \lambda = 0.5, \mathbf{v} = \begin{pmatrix}1 \\ -1\end{pmatrix}.\)
• \(P = \begin{pmatrix}3 & 1 \\ 2 & -1\end{pmatrix},\; D = \begin{pmatrix}1 & 0 \\ 0 & 0.5\end{pmatrix},\; P^{-1} = \frac{1}{-5}\begin{pmatrix}-1 & -1 \\ -2 & 3\end{pmatrix}\)
• \(\text{(a) } T^3 = PD^3P^{-1} \text{ with } D^3 = \begin{pmatrix}1 & 0 \\ 0 & 0.125\end{pmatrix}.\)
• \(\text{From the calculator: } T^3 = \begin{pmatrix}0.65 & 0.525 \\ 0.35 & 0.475\end{pmatrix}\)
• \(T^3\begin{pmatrix}5000 \\ 5000\end{pmatrix} = \begin{pmatrix}5875 \\ 4125\end{pmatrix}\)
• After 3 years: 5875 in the town and 4125 in the countryside.
• \(\text{(b) As } n \to \infty,\; 0.5^n \to 0, \text{ so } D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}.\)
• \(T^n \to \begin{pmatrix}3 & 1 \\ 2 & -1\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\tfrac{1}{5}\begin{pmatrix}-1 & -1 \\ -2 & 3\end{pmatrix} = \begin{pmatrix}0.6 & 0.6 \\ 0.4 & 0.4\end{pmatrix}\)
• \(\begin{pmatrix}0.6 & 0.6 \\ 0.4 & 0.4\end{pmatrix}\begin{pmatrix}5000 \\ 5000\end{pmatrix} = \begin{pmatrix}6000 \\ 4000\end{pmatrix}\)
• In the long term: 6000 in the town and 4000 in the countryside.

Each year every adult beetle produces 2.2 young, 20% of the young survive to become adults, and 70% of the adults survive. Let \(y_n\) and \(a_n\) be the numbers of young and adult beetles in year \(n\).

1. (a) Write down the matrix \(M\) with \(\begin{pmatrix}y_{n+1} \\ a_{n+1}\end{pmatrix} = M\begin{pmatrix}y_n \\ a_n\end{pmatrix}\).

   • New young come from adults: \(y_{n+1} = 2.2\,a_n\). New adults: \(a_{n+1} = 0.2\,y_n + 0.7\,a_n\)
   • \(M = \begin{pmatrix}0 & 2.2 \\ 0.2 & 0.7\end{pmatrix}\)

2. (b) Find the eigenvalues of \(M\).

   • \(\lambda^2 - 0.7\lambda - 0.44 = 0\)
   • \(\lambda = 1.1 \text{ or } -0.4\)

3. (c) Find the long-term percentage increase in the population each year, and the long-term percentage of the beetles that are young.

   • The larger eigenvalue \(1.1\) gives the long-term multiplier: an increase of 10%.
   • Eigenvector: \(\begin{pmatrix}2 \\ 1\end{pmatrix}\)
   • Share that are young: \(\dfrac{2}{2+1} = 0.667\), which is 66.7%.

Explain why \(M^n = PD^nP^{-1}\) when \(M = PDP^{-1}\).

• \(M^n = (PDP^{-1})(PDP^{-1})\cdots(PDP^{-1}) \quad (n \text{ brackets})\)
• \(\text{Each neighbouring pair is } P^{-1}P = I \text{, and } DI = D, \text{ so the middle terms cancel.}\)
• \(M^n = PD^nP^{-1}\)

The eigenvalues of a matrix \(M\) are 1 and 0.4. Explain why \(M^n\) settles to a fixed matrix as \(n\) grows. What would happen if they were 1 and 1.4?

• \(M = PDP^{-1} \text{ with } D = \begin{pmatrix}1 & 0 \\ 0 & 0.4\end{pmatrix}, \text{ so } D^n = \begin{pmatrix}1 & 0 \\ 0 & 0.4^n\end{pmatrix}.\)
• \(0.4^n \to 0 \text{ as } n \to \infty, \text{ so } D^n \to \begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix} \text{ and } M^n \to P\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}P^{-1}.\)
• \(\text{With eigenvalues 1 and 1.4, } 1.4^n \to \infty \text{, so the entries of } M^n \text{ grow without limit and do not settle.}\)