AI HL · AHL 5
0 of 63 skills done
Diagonalisation and powers of matrices
AHL 5.7
a^n
Screenshot and video to come
The unit square is transformed by M, then by M again: n times in all, which is Mⁿ. Raise n one step at a time. Before each step, say where the square will go next.
\(M = \begin{pmatrix}4 & 1 \\ 2 & 3\end{pmatrix}\) has eigenvalues 5 and 2 with eigenvectors \(\begin{pmatrix}1 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -2\end{pmatrix}\). Write \(M\) in the form \(PDP^{-1}\).
Have a go first. Then open the working one line at a time.
\(P = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix}\)
The eigenvectors are the columns.
\(D = \begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix}\)
5 goes with the first column, 2 with the second.
\(M = \begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix}\begin{pmatrix}5 & 0 \\ 0 & 2\end{pmatrix}\begin{pmatrix}1 & 1 \\ 1 & -2\end{pmatrix}^{-1}\)
Check it on the calculator.
For the same matrix, find \(M^n\) in terms of \(n\), and hence find \(M^3\).
Have a go first. Then open the working one line at a time.
\(M^n = PD^nP^{-1}\)
Only \(D\) is raised to the power.
\(P^{-1} = \frac{1}{-3}\begin{pmatrix}-2 & -1 \\ -1 & 1\end{pmatrix} = \frac{1}{3}\begin{pmatrix}2 & 1 \\ 1 & -1\end{pmatrix}\)
\(\det P = -3\).
\(M^n = \frac{1}{3}\begin{pmatrix}2(5^n) + 2^n & 5^n - 2^n \\ 2(5^n) - 2(2^n) & 5^n + 2(2^n)\end{pmatrix}\)
\(P\) times \(D^n\) first, then times \(P^{-1}\).
\(M^3 = \frac{1}{3}\begin{pmatrix}258 & 117 \\ 234 & 141\end{pmatrix} = \begin{pmatrix}86 & 39 \\ 78 & 47\end{pmatrix}\)
\(5^3 = 125\) and \(2^3 = 8\). Check with a^3.
\(T = \begin{pmatrix}0.9 & 0.3 \\ 0.1 & 0.7\end{pmatrix}\) has eigenvalues 1 and 0.6 with eigenvectors \(\begin{pmatrix}3 \\ 1\end{pmatrix}\) and \(\begin{pmatrix}1 \\ -1\end{pmatrix}\). Find the matrix that \(T^n\) approaches as \(n\) becomes large.
Have a go first. Then open the working one line at a time.
\(T^n = P\begin{pmatrix}1 & 0 \\ 0 & 0.6^n\end{pmatrix}P^{-1}\) with \(P = \begin{pmatrix}3 & 1 \\ 1 & -1\end{pmatrix}\)
The eigenvectors are the columns of \(P\).
\(0.6^n \to 0\)
Only the eigenvalue 1 is left.
\(\begin{pmatrix}3 & 1 \\ 1 & -1\end{pmatrix}\begin{pmatrix}1 & 0 \\ 0 & 0\end{pmatrix}\cdot\frac{1}{4}\begin{pmatrix}1 & 1 \\ 1 & -3\end{pmatrix} = \begin{pmatrix}0.75 & 0.75 \\ 0.25 & 0.25\end{pmatrix}\)
\(P^{-1} = \frac{1}{4}\begin{pmatrix}1 & 1 \\ 1 & -3\end{pmatrix}\).
T^50 on the calculator
Both columns are the same.
The numbers of young birds (\(y\)) and adult birds (\(a\)) on an island change each year by \(\begin{pmatrix}y_{n+1} \\ a_{n+1}\end{pmatrix} = \begin{pmatrix}0.5 & 1.2 \\ 0.4 & 0.7\end{pmatrix}\begin{pmatrix}y_n \\ a_n\end{pmatrix}\). At the start there are 140 young and 70 adults. The matrix has eigenvalues 1.3 and −0.1 with eigenvectors \(\begin{pmatrix}3 \\ 2\end{pmatrix}\) and \(\begin{pmatrix}2 \\ -1\end{pmatrix}\). Find the numbers after \(n\) years, and describe what happens in the long term.
Have a go first. Then open the working one line at a time.
\(M^n\begin{pmatrix}140 \\ 70\end{pmatrix}\), with \(M^n = PD^nP^{-1}\)
Use the matrix \(n\) times on the start.
\(P = \begin{pmatrix}3 & 2 \\ 2 & -1\end{pmatrix},\quad D^n = \begin{pmatrix}1.3^n & 0 \\ 0 & (-0.1)^n\end{pmatrix}\)
The eigenvectors are the columns of \(P\). The eigenvalues go in the same order.
\(\frac{1}{7}\begin{pmatrix}1 & 2 \\ 2 & -3\end{pmatrix}\begin{pmatrix}140 \\ 70\end{pmatrix} = \begin{pmatrix}40 \\ 10\end{pmatrix}\)
\(\det P = -7\). Work from the right: \(P^{-1}\) meets the start first.
\(\begin{pmatrix}3 & 2 \\ 2 & -1\end{pmatrix}\begin{pmatrix}40(1.3^n) \\ 10(-0.1)^n\end{pmatrix} = \begin{pmatrix}120(1.3^n) + 20(-0.1)^n \\ 80(1.3^n) - 10(-0.1)^n\end{pmatrix}\)
Check \(n = 1\): 154 young and 105 adults, the same as \(M\) times the start.
\((-0.1)^n \to 0\), so the numbers approach \(1.3^n\begin{pmatrix}120 \\ 80\end{pmatrix}\)
Young to adults settles at 3 : 2, the eigenvector for 1.3. Both numbers rise by 30% each year.
A matrix \(M\) has eigenvalue 3 with eigenvector \(\begin{pmatrix}2 \\ 1\end{pmatrix}\) and eigenvalue \(-1\) with eigenvector \(\begin{pmatrix}1 \\ 1\end{pmatrix}\). Find \(M\).
Have a go first. Then open the working one line at a time.
\(P = \begin{pmatrix}2 & 1 \\ 1 & 1\end{pmatrix},\quad D = \begin{pmatrix}3 & 0 \\ 0 & -1\end{pmatrix}\)
The eigenvectors are the columns of \(P\). The eigenvalues go on the diagonal of \(D\) in the same order.
\(\det P = 2(1) - 1(1) = 1,\quad P^{-1} = \begin{pmatrix}1 & -1 \\ -1 & 2\end{pmatrix}\)
Swap the 2 and the 1 on the leading diagonal, change the signs of the other two.
\(M = PDP^{-1} = \begin{pmatrix}6 & -1 \\ 3 & -1\end{pmatrix}\begin{pmatrix}1 & -1 \\ -1 & 2\end{pmatrix} = \begin{pmatrix}7 & -8 \\ 4 & -5\end{pmatrix}\)
Work from the left. \(PD\) first.
Trace = 2 = 3 + (−1)
The eigenvalues add to the trace.
Each year, every adult fish produces 1.5 young, 40% 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 fish in year \(n\). Find the long-term percentage increase in the population each year, and the long-term percentage of the fish that are young.
Have a go first. Then open the working one line at a time.
\(\begin{pmatrix}y_{n+1} \\ a_{n+1}\end{pmatrix} = \begin{pmatrix}0 & 1.5 \\ 0.4 & 0.7\end{pmatrix}\begin{pmatrix}y_n \\ a_n\end{pmatrix}\)
New young come only from adults. New adults come from surviving young and surviving adults.
\(\lambda^2 - 0.7\lambda - 0.6 = 0,\quad \lambda = 1.2 \text{ or } -0.5\)
From the calculator. The eigenvalue with the larger size, 1.2, takes over, because \((-0.5)^n\) fades away.
Each year the population is multiplied by 1.2 in the long term: an increase of 20%.
The long-term multiplier is the larger eigenvalue.
\(1.5a = 1.2y\)
So \(y : a = 5 : 4\). Choose \(\begin{pmatrix}5 \\ 4\end{pmatrix}\).
\(\dfrac{5}{5+4} = 0.556\)
In the long term 55.6% of the fish are young, whatever the start (as long as it is not made of the other eigenvector).