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Why $A^{k} = PD^{k}P^{-1}$, how a linear recurrence becomes a matrix power, and how the sizes of the eigenvalues decide whether repeated multiplication decays, settles or runs away.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you will be able to raise a diagonal matrix to a power and use $A^{k} = PD^{k}P^{-1}$ to raise a diagonalisable one, expand a starting vector in the eigenbasis and describe its whole trajectory as independent geometric sequences, convert a linear recurrence into a matrix and read its growth rate off the dominant eigenvalue, classify long-run behaviour by comparing each eigenvalue's absolute value with one, and find the steady state of a two-state transition model as an eigenvector for eigenvalue one.
A diagonalisable $A$ is $PDP^{-1}$, with an eigenbasis in the columns of $P$ and the matching eigenvalues down $D$. That factorisation has so far been a way of describing a matrix. This lesson is what it is for: everything below is one identity and its consequences.
The dominant eigenvalue is the one of largest absolute value. A linear recurrence such as $u_{n+1} = au_n + bu_{n-1}$ becomes a matrix acting on a vector of consecutive terms. A square matrix whose columns are non-negative and sum to $1$ is a stochastic or transition matrix; a vector it leaves unchanged is a steady state, which is an eigenvector for $\lambda = 1$. A matrix is stable when every eigenvalue has absolute value below $1$, so that repeated application drives every vector to zero.
Write $A = PDP^{-1}$ and square it:
$$A^{2} = PDP^{-1} \cdot PDP^{-1} = PD(P^{-1}P)DP^{-1} = PD^{2}P^{-1},$$
and the same cancellation repeated gives
$$A^{k} = PD^{k}P^{-1} \quad \text{for every } k \ge 1.$$
$D^{k}$ is free — a diagonal matrix to a power is each diagonal entry to that power, since the rows never mix — so a hundredth power of an $n \times n$ matrix costs $n$ numerical powers and two matrix products, instead of ninety-nine matrix products. That is the computational point, and it is a large one.
The conceptual point is larger. Read $A^{k}x$ from the right. $P^{-1}x$ re-expresses $x$ in the eigenbasis as coordinates $c_1, \dots, c_n$, so that
$$x = c_1v_1 + \dots + c_nv_n \quad \Longrightarrow \quad A^{k}x = c_1\lambda_1^{k}v_1 + \dots + c_n\lambda_n^{k}v_n.$$
The coordinates never interact. Each one is simply multiplied by its own $\lambda^{k}$, so the whole future of the sequence $x, Ax, A^{2}x, \dots$ is $n$ independent geometric sequences, and their behaviour is decided by one comparison each:
| $|\lambda|$ | what that coordinate does |
|---|---|
| less than $1$ | decays to zero |
| equal to $1$ | stays exactly as it was |
| greater than $1$ | grows without bound |
After enough steps only the largest $|\lambda|$ with a non-zero coordinate is worth anything, so $A^{k}x$ lines up with the dominant eigenvector — whatever $x$ was.
Another way: picture
Draw the two eigenvector directions as a skewed pair of axes. A starting arrow has a component along each. Applying $A$ once stretches the first component by $\lambda_1$ and the second by $\lambda_2$, and applying it again does the same to the result. If $\lambda_1 = 3$ and $\lambda_2 = \tfrac{1}{2}$, then after ten steps the first component has been multiplied by about $59{,}000$ and the second by about $\tfrac{1}{1000}$: the arrow has swung round until it is all but parallel to the first eigenvector, and it does so from almost any starting direction.
Another way: steps
The rule $u_{n+1} = u_n + u_{n-1}$ needs two terms of history, so make the state the pair of consecutive terms. Writing $x_n = (u_n, u_{n-1})$,
$$x_{n+1} = \begin{pmatrix} u_{n+1} \\ u_n \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} u_n \\ u_{n-1} \end{pmatrix} = Fx_n,$$
so $x_n = F^{n-1}x_1$ and the whole sequence is one matrix power. $F$ has trace $1$ and determinant $-1$, so its characteristic polynomial is $t^{2} - t - 1$ and its eigenvalues are $\varphi = \tfrac{1 + \sqrt{5}}{2} \approx 1.618$ and $\tfrac{1 - \sqrt{5}}{2} \approx -0.618$.
Two things fall straight out. The closed form for the Fibonacci numbers is the eigenbasis expansion written out — a combination of $\varphi^{n}$ and $(1 - \varphi)^{n}$, which is Binet's formula. And the ratio of consecutive terms tends to $\varphi$ for every starting pair, because $|1 - \varphi| < 1$ kills the second coordinate whatever it was. The golden ratio is not a fact about that particular sequence; it is the dominant eigenvalue of that particular matrix.
The same conversion handles any linear recurrence of any order: a rule needing $m$ terms of history becomes an $m \times m$ companion matrix, and the growth rate of the sequence is its dominant eigenvalue.
A transition matrix moves a population between states: column $j$ says how what is in state $j$ is distributed next step, so its entries are non-negative and sum to $1$. Because the columns sum to $1$, the total population never changes, and it can be shown that $\lambda = 1$ is always an eigenvalue. Its eigenvector is the steady state — the distribution the step leaves alone.
If every other eigenvalue has absolute value strictly below $1$, then the eigenbasis expansion says exactly what happens: the coordinate along the steady state is frozen, every other coordinate decays, and the population converges to the steady state from any starting distribution at all. A model like that forgets its initial conditions, which is precisely why a long-run prediction can be made without knowing them.
Finding the steady state does not need the machinery. It is the solution of $(A - I)v = 0$, and for two states it is the balance condition written directly: as many leave a state each step as enter it. Scale the answer so its entries sum to the total, and the arithmetic is over.
The same eigenvalue comparison classifies dynamical systems generally. Every $|\lambda| < 1$ is a stable system, returning to equilibrium after a disturbance; one $|\lambda| > 1$ is unstable, and a disturbance grows; $|\lambda| = 1$ exactly is the borderline that neither recovers nor blows up, and it is where a complex pair of modulus $1$ produces a system that oscillates for ever.
$A^{k} \ne P^{k}D^{k}(P^{-1})^{k}$. The cancellation is the whole trick, and it works only in the order $PDP^{-1}$. Powers of $P$ never appear.
Raising the entries of $A$ to the power. $A^{2}$ is a matrix product, not an entrywise square. $D^{2}$ is the entrywise square, and only because $D$ is diagonal.
Forgetting that the eigenvectors do not move. $A^{k}$, $A + cI$ and any polynomial in $A$ all have the same eigenvectors as $A$. Only the eigenvalues change, and they change in the obvious way. This is often the quickest route to an answer.
Assuming the largest eigenvalue always wins. It wins if the starting vector has a non-zero coordinate along its eigenvector. A vector that happens to lie exactly in another eigenspace stays there for ever — a set of measure zero, but the reason the qualification is in every careful statement.
Confusing $|\lambda|$ with $\lambda$. $\lambda = -2$ grows, and its coordinate flips sign at every step. It is the absolute value that decides growth and the sign that decides oscillation.
$A = \begin{pmatrix} 4 & 1 \\ 2 & 3 \end{pmatrix} = PDP^{-1}$ with $P = \begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix}$ and $D = \operatorname{diag}(5, 2)$.
The diagonalisation from the last lesson.
$A^{10} = PD^{10}P^{-1}$ with $D^{10} = \operatorname{diag}(5^{10}, 2^{10}) = \operatorname{diag}(9765625, 1024)$.
Two numerical powers, not ten matrix products.
$5^{10}$ is about $9500$ times $2^{10}$, so $A^{10}x$ is almost exactly a multiple of $(1, 1)$ for any $x$ with a component in that direction.
The dominant eigenvalue has taken over.
Each year $20\%$ of city dwellers move to the suburbs and $10\%$ of suburb dwellers move to the city, in a region of $3$ million people.
Two states, fixed total.
Steady state: the flows balance, so $0.2c = 0.1s$, giving $s = 2c$. With $c + s = 3$ million, $c = 1$ million and $s = 2$ million.
An eigenvector for $\lambda = 1$.
The other eigenvalue is $1 - 0.2 - 0.1 = 0.7$, so any departure from that split shrinks by $30\%$ a year and the region converges to it from any starting split whatever.
The second eigenvalue sets the speed, not the destination.
Expand $x$ in the eigenbasis: $x = c_1v_1 + c_2v_2$, so $A^{k}x = c_1 \cdot 1^{k}v_1 + c_2(\tfrac{1}{5})^{k}v_2$.
Two independent geometric sequences.
$1^{k} = 1$ for every $k$, and $(\tfrac{1}{5})^{k} \to 0$.
Compare each with one.
So $A^{k}x \to c_1v_1$: it settles on the eigenvector for $1$, scaled by however much of it $x$ contained. Nothing else survives.
For $D = \begin{pmatrix} 3 & 0 \\ 0 & 2 \end{pmatrix}$, write $D^{3}$.
This task has no paper form; do it on a device.
A sequence has $u_0 = 7$, $u_1 = 7$ and $u_{n+1} = u_n + u_{n-1}$. The matrix $F = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$ sends the column $(u_n, u_{n-1})$ to $(u_{n+1}, u_n)$. Find $u_4$.
Answer:
A matrix has eigenvalues $5$ and $2$. Fill in the diagonal entries of $D^{2}$ and $D^{3}$.
| Entry of $D^{2}$ | Entry of $D^{3}$ | |
|---|---|---|
| $\lambda = 5$ | ||
| $\lambda = 2$ |
A non-zero vector $x$ is multiplied by $A$ again and again, giving $x$, $Ax$, $A^{2}x$, and so on. Given that the eigenvalues are $1$ and $\tfrac{1}{4}$, and the starting vector has a non-zero component along the eigenvector for $1$, what happens to $A^{k}x$ as $k$ grows?
$A = \begin{pmatrix} 10 & -2 \\ 24 & -4 \end{pmatrix}$ has eigenvector $(1, 3)$ for $\lambda = 4$ and $(1, 4)$ for $\lambda = 2$. Write $A^{2}$.
This task has no paper form; do it on a device.
Every month $\tfrac{4}{10}$ of a town's cyclists switch to the bus, and $\tfrac{1}{10}$ of its bus passengers switch to cycling. The town has $15$ commuters in all, and nobody joins or leaves. In the long run, how many cycle and how many take the bus? Write the answer as a column with the cyclists on top.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $D = \begin{pmatrix} 3 & 0 \\ 0 & 4 \end{pmatrix}$, write $D^{3}$.
This task has no paper form; do it on a device.
You can compute a power through the diagonalisation, turn a recurrence into a matrix, and say from the eigenvalues whether repeated multiplication decays, settles or grows. Say in your own words why the eigenvectors of $A^{k}$ are the same as those of $A$. Next: lengths and angles, and what orthogonality adds to all of this.
10. Your turn: $A$ has eigenvalues $1$ and $\tfrac{1}{5}$. What is $A^{k}x$ doing for large $k$?, step 3
The destination depends on $x$ only through $c_1$.