Multiply a vector by a matrix, multiply the answer by the same matrix, and keep going: $x_0, Ax_0, A^2x_0, \ldots$ A population counted year after year, a balance month after month, the Fibonacci numbers one after another: each is a matrix applied over and over. This page computes $A^kx_0$ without multiplying $k$ times, by splitting $x_0$ along eigenvectors, and then reads off where the process heads in the long run.
Pick a point, multiply it by a matrix, then multiply the result by the same matrix, and keep going. The dots trace out a path. Where does it go? Off to infinity, in to $0$, flipping from side to side? The answer is hiding in the matrix's eigenvectors, and this page reads it off with pictures.
A true-or-false question on Eigenvalues & Eigenvectors said: if $Av = \lambda v$, then $A^2v = \lambda^2v$. Do it $k$ times and $A^kv = \lambda^kv$. On an eigenvector, a matrix power is just a number power. This whole page grows out of that one line.
The dots in that figure drift toward one particular line and then march straight out along it. By the end of the page you'll be able to predict that line, and the speed of the march, before drawing a single dot.
The Fibonacci numbers $0, 1, 1, 2, 3, 5, 8, 13, \ldots$ follow a rule: each one is the sum of the two before it. Rules like that get a name.
A linear recurrence of order $m$ is a rule $a_{k+m} = p_1a_{k+m-1} + p_2a_{k+m-2} + \cdots + p_ma_k$ (fixed numbers $p_1, \ldots, p_m$, with $p_m \neq 0$) that builds each term of a sequence $(a_k)$ from the $m$ terms before it. The starting values $a_0, \ldots, a_{m-1}$ then decide the whole sequence.
A linear recurrence builds each term of a sequence $(a_k)$ from the terms just before it. Order $2$ uses two: $a_{k+2} = p_1a_{k+1} + p_2a_k$ (with $p_2 \neq 0$). Order $3$ uses three: $a_{k+3} = p_1a_{k+2} + p_2a_{k+1} + p_3a_k$ (with $p_3 \neq 0$). The starting values ($a_0, a_1$, or $a_0, a_1, a_2$) then decide the whole sequence.
A recurrence of order $2$ has to remember the last two terms, so stack them into a vector with the newest on top: $x_k = (a_{k+1}, a_k)$. Then one step of the rule is one multiplication by a matrix.
Write the Fibonacci rule $F_{k+2} = F_{k+1} + F_k$, with $F_0 = 0$ and $F_1 = 1$, as $x_{k+1} = Ax_k$.
Take $x_k = (F_{k+1}, F_k)$. The next vector is
$$x_{k+1} = \begin{pmatrix} F_{k+2} \\ F_{k+1} \end{pmatrix} = \begin{pmatrix} F_{k+1} + F_k \\ F_{k+1} \end{pmatrix} = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} F_{k+1} \\ F_k \end{pmatrix}.$$The top row builds the newest number; the bottom row just copies $F_{k+1}$ down one slot. So $A = \begin{pmatrix} 1 & 1 \\ 1 & 0 \end{pmatrix}$, $x_0 = (1, 0)$, and $x_k = A^kx_0$: the $k$th Fibonacci number is the bottom entry of $A^k(1, 0)$.
Picture $x_k$ as a point in the plane: $(1, 0), (1, 1), (2, 1), (3, 2), (5, 3), \ldots$ Each hop is one multiplication by $A$.
Fibonacci's rabbits never die, never get sick, and always come in exactly one boy-girl pair. Most optimistic population model ever written.
The same trick works for any order: the first row holds the coefficients, and $1$s just below the diagonal copy each term down one slot. The stacking is a choice, though, and the matrix has to match it.
Keep the newest term on top. If you stack $x_k = (a_k, a_{k+1})$ instead, the matrix turns upside down: $\begin{pmatrix} 0 & 1 \\ p_2 & p_1 \end{pmatrix}$ for order $2$. Either is right, as long as the matrix goes with the stacking you picked. (Other books call these companion matrices, and some put the coefficients down the last column instead.)
The eigenvalues of that matrix are where the action is, and you don't even have to expand a determinant to get them: the recurrence hands you its characteristic polynomial.
Let $A$ be the $m \times m$ matrix of the recurrence $a_{k+m} = p_1a_{k+m-1} + \cdots + p_ma_k$, with $x_k = (a_{k+m-1}, \ldots, a_{k+1}, a_k)$: the coefficients across its first row and $1$s just below its diagonal. Then
$$\det(A - tI) = (-1)^m\left(t^m - p_1t^{m-1} - \cdots - p_{m-1}t - p_m\right).$$So the eigenvalues of $A$ are the roots of the recurrence's polynomial $t^m - p_1t^{m-1} - \cdots - p_m$, and for each root $r$, the vector $(r^{m-1}, \ldots, r, 1)$ is an eigenvector.
Expand $\det(A - tI)$ along the first row, whose entries are $p_1 - t, p_2, \ldots, p_m$ (cofactor expansion). Delete row $1$ and column $j$. The rows $2, \ldots, j$ and columns $1, \ldots, j-1$ that remain form a lower triangular block with $1$s on its diagonal (determinant $1$), the rows and columns after $j$ form a lower triangular block with $-t$ on its diagonal (determinant $(-t)^{m-j}$), and the block above it is all $0$s. So that minor is $(-t)^{m-j}$, and
$$\det(A - tI) = (p_1 - t)(-t)^{m-1} + \sum_{j=2}^{m} (-1)^{1+j}p_j(-t)^{m-j} = (-1)^m t^m + (-1)^{m+1}\sum_{j=1}^{m} p_jt^{m-j},$$since $(-1)^{1+j}(-1)^{m-j} = (-1)^{m+1}$. Factoring out $(-1)^m$ gives the formula. For the eigenvector: if $r$ is a root, then $A(r^{m-1}, \ldots, r, 1)$ has first entry $p_1r^{m-1} + \cdots + p_m = r^m$, and its other entries are the copied $r^{m-1}, \ldots, r$. That is $r(r^{m-1}, \ldots, r, 1)$. $\blacksquare$
The matrix $\begin{pmatrix} p_1 & p_2 \\ 1 & 0 \end{pmatrix}$ of $a_{k+2} = p_1a_{k+1} + p_2a_k$ has as eigenvalues the roots of the recurrence's polynomial $t^2 - p_1t - p_2$, and $(r, 1)$ is an eigenvector for each root $r$. For order $3$, the matrix $\begin{pmatrix} p_1 & p_2 & p_3 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$ has as eigenvalues the roots of $t^3 - p_1t^2 - p_2t - p_3$, with eigenvector $(r^2, r, 1)$ for each root $r$.
For order $2$, $\det\begin{pmatrix} p_1 - t & p_2 \\ 1 & -t \end{pmatrix} = (p_1 - t)(-t) - p_2 = t^2 - p_1t - p_2$. And $\begin{pmatrix} p_1 & p_2 \\ 1 & 0 \end{pmatrix}\begin{pmatrix} r \\ 1 \end{pmatrix} = \begin{pmatrix} p_1r + p_2 \\ r \end{pmatrix}$, which is $\begin{pmatrix} r^2 \\ r \end{pmatrix} = r\begin{pmatrix} r \\ 1 \end{pmatrix}$ exactly when $r^2 = p_1r + p_2$, that is, when $r$ is a root. The $3 \times 3$ is the same check one size up: the copying rows shift $(r^2, r, 1)$ down to $(\,\cdot\,, r^2, r)$, and the top row gives $p_1r^2 + p_2r + p_3 = r^3$.
The full proof is in the n version.
Here's the example the rest of the page keeps coming back to.
Find the matrix of $a_{k+2} = a_{k+1} + 2a_k$, its eigenvalues, and an eigenvector for each.
With $x_k = (a_{k+1}, a_k)$, the matrix is $A = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}$. Its polynomial comes straight off the rule: $t^2 - t - 2 = (t - 2)(t + 1)$, so the eigenvalues are $2$ and $-1$, with eigenvectors $(2, 1)$ and $(-1, 1)$.
Check one: $A(2, 1) = (2 + 2, \ 2) = (4, 2) = 2(2, 1)$. Good.
In the plane, $A$ stretches the line through $(2, 1)$ by $2$ and flips the line through $(-1, 1)$. The next section's picture runs on exactly these two lines.
Your turn: turn a rule into a matrix. In free response, type the matrix row by row with the rows split by semicolons, like 2, 5; 1, 0.
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Now the payoff. Computing $A^{10}x_0$ by brute force takes ten multiplications. But on an eigenvector, $A^k$ is just $\lambda^k$. So split $x_0$ into eigenvectors first, and let each piece grow on its own.
Let $A$ be an $n \times n$ matrix with eigenvectors $v_1, \ldots, v_n$ that form a basis of $\R^n$, with $Av_i = \lambda_iv_i$. If $x_0 = c_1v_1 + \cdots + c_nv_n$, then for every $k \ge 0$,
$$A^kx_0 = c_1\lambda_1^kv_1 + c_2\lambda_2^kv_2 + \cdots + c_n\lambda_n^kv_n.$$First, $A^kv_i = \lambda_i^kv_i$ for every $k$, by induction: it holds for $k = 0$, and if $A^kv_i = \lambda_i^kv_i$, then $A^{k+1}v_i = A(\lambda_i^kv_i) = \lambda_i^kAv_i = \lambda_i^{k+1}v_i$. Then, since $A^k$ is a matrix, it respects sums and scalar multiples: $A^kx_0 = c_1A^kv_1 + \cdots + c_nA^kv_n = c_1\lambda_1^kv_1 + \cdots + c_n\lambda_n^kv_n$. $\blacksquare$
Let $A$ be a $2 \times 2$ matrix with eigenvectors $v_1, v_2$ on two different lines (or a $3 \times 3$ matrix with eigenvectors $v_1, v_2, v_3$ not all in one plane), with $Av_i = \lambda_iv_i$. If $x_0 = c_1v_1 + c_2v_2$ (or $c_1v_1 + c_2v_2 + c_3v_3$), then for every $k$,
$$A^kx_0 = c_1\lambda_1^kv_1 + c_2\lambda_2^kv_2 \quad (+\ c_3\lambda_3^kv_3).$$This is a visual argument, not a formal proof. Split $x_0$ into legs along the eigenvectors, laid tip to tail. $A$ only stretches each leg along its own line, by its own $\lambda$, and it carries a tip-to-tail chain to the chain of the images, so the tip of the new legs is $Ax_0$. After $k$ steps, leg $i$ has been stretched $k$ times: by $\lambda_i^k$. Here it is for the running example's matrix $A = \begin{pmatrix} 1 & 2 \\ 1 & 0 \end{pmatrix}$, starting from $x_0 = (7, -1) = 2v_1 - 3v_2$.
The full proof is in the n version.
That is the same fact as $A^k = QD^kQ^{-1}$ on Diagonalization, read one vector at a time: $Q$'s columns are the $v_i$, $D^k$ holds the $\lambda_i^k$, and splitting $x_0$ is solving $Qc = x_0$. You never need $Q^{-1}$ or the whole of $A^k$, only the numbers $c_i$.
The sequence $a_{k+2} = a_{k+1} + 2a_k$ starts $a_0 = 1$, $a_1 = 5$. Find a formula for $a_k$.
The start is $x_0 = (a_1, a_0) = (5, 1)$. Split it along the eigenvectors $(2, 1)$ (for $2$) and $(-1, 1)$ (for $-1$): $c_1(2, 1) + c_2(-1, 1) = (5, 1)$ means $2c_1 - c_2 = 5$ and $c_1 + c_2 = 1$, so $c_1 = 2$ and $c_2 = -1$. By the theorem,
$$x_k = A^kx_0 = 2 \cdot 2^k\begin{pmatrix} 2 \\ 1 \end{pmatrix} - (-1)^k\begin{pmatrix} -1 \\ 1 \end{pmatrix}.$$The bottom entry is $a_k$, so $a_k = 2 \cdot 2^k - (-1)^k$. Check: $a_2 = 8 - 1 = 7$, and the rule gives $5 + 2 \cdot 1 = 7$; $a_3 = 16 + 1 = 17 = 7 + 2 \cdot 5$.
The formula is two simple sequences stacked: one doubling, one flipping between $-1$ and $1$.
Only the bottom entry mattered, and every eigenvector of a recurrence's matrix ends in a $1$. So the bottom entry of $c_ir_i^k(\ldots, r_i, 1)$ is just $c_ir_i^k$, and the formula always has the same shape.
If the polynomial $t^m - p_1t^{m-1} - \cdots - p_m$ of a linear recurrence has $m$ different roots $r_1, \ldots, r_m$, then every sequence that follows the recurrence is
$$a_k = c_1r_1^k + c_2r_2^k + \cdots + c_mr_m^k$$for some numbers $c_1, \ldots, c_m$, and the starting values decide them.
Let $x_k = (a_{k+m-1}, \ldots, a_k)$, so $x_k = A^kx_0$ for the recurrence's matrix $A$. By the theorem above, $w_i = (r_i^{m-1}, \ldots, r_i, 1)$ is an eigenvector for $r_i$. The $r_i$ are different, so the $w_i$ are linearly independent, and $m$ of them form a basis of $\R^m$. Write $x_0 = c_1w_1 + \cdots + c_mw_m$. By the powers theorem, $x_k = c_1r_1^kw_1 + \cdots + c_mr_m^kw_m$, and its last entry is $a_k = c_1r_1^k + \cdots + c_mr_m^k$. The $c_i$ come from $x_0$, which is the starting values. $\blacksquare$
This is a visual argument, not a formal proof. It's the legs picture above: the start splits into one leg per eigenvector, each leg grows like $r_i^k$, and every eigenvector of a recurrence's matrix ends in $1$, so the bottom entry of the tip is $c_1r_1^k + c_2r_2^k\ (+\ c_3r_3^k)$. Different roots give eigenvectors on different lines, which is what lets the start split at all.
The full proof is in the n version.
So in practice you can skip the eigenvectors entirely.
Shortcut: find the roots, write $a_k = c_1r_1^k + c_2r_2^k$, and plug in the starting values to get two equations for $c_1$ and $c_2$. Watch the indices: if you're given $a_1$ and $a_2$, plug in $k = 1$ and $k = 2$, not $0$ and $1$. (A repeated root needs a different form, $c_1r^k + c_2kr^k$, which this page doesn't cover.)
Here's the shortcut on its own.
Solve $a_{k+2} = 5a_{k+1} - 6a_k$ with $a_0 = 1$ and $a_1 = 1$.
The polynomial is $t^2 - 5t + 6 = (t - 2)(t - 3)$, so $a_k = c_1 \cdot 2^k + c_2 \cdot 3^k$. Then $k = 0$ gives $c_1 + c_2 = 1$ and $k = 1$ gives $2c_1 + 3c_2 = 1$, so $c_2 = -1$ and $c_1 = 2$:
$$a_k = 2 \cdot 2^k - 3^k.$$Check: the rule gives $a_2 = 5 - 6 = -1$, and the formula gives $8 - 9 = -1$. After that the terms go negative for good, since the $3^k$ part outgrows the $2^k$ part.
The most famous closed form of all comes out of the Fibonacci matrix, square roots and all.
Find a formula for the Fibonacci numbers.
The polynomial is $t^2 - t - 1$, with roots $\varphi = \frac{1 + \sqrt 5}{2} \approx 1.618$ (the golden ratio) and $\overline{\varphi} = \frac{1 - \sqrt 5}{2} \approx -0.618$. So $F_k = c_1\varphi^k + c_2\overline{\varphi}^k$. From $F_0 = 0$, $c_2 = -c_1$; from $F_1 = 1$, $c_1(\varphi - \overline{\varphi}) = 1$, and $\varphi - \overline{\varphi} = \sqrt 5$. So $c_1 = \frac{1}{\sqrt 5}$ and
$$F_k = \frac{\varphi^k - \overline{\varphi}^k}{\sqrt 5}.$$This is Binet's formula. Since $|\overline{\varphi}| < 1$, the second term is always less than $\frac12$ in size, so $F_k$ is just $\varphi^k/\sqrt 5$ rounded to the nearest whole number: $\varphi^{10}/\sqrt 5 \approx 55.004$, and $F_{10} = 55$.
Here are both halves of the formula side by side.
A formula for whole numbers, built out of $\sqrt 5$, and every irrational crumb cancels perfectly. Every single time. I don't trust it. I checked $F_{10}$ by hand anyway.
Your turn. In free response, type the formula with $k$ in it, like 2*3^k - (-1)^k.
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In the running example, $a_k = 2 \cdot 2^k - (-1)^k$: the first part doubles forever, and the second just flips between $-1$ and $1$. After a few steps the doubling part is all you can see, and $a_{k+1}/a_k$ gets closer and closer to $2$. In general the eigenvalue of biggest size takes over, and it gets a name.
An eigenvalue $\lambda_1$ of a square matrix $A$ is dominant if $|\lambda_1| > |\lambda|$ for every other eigenvalue $\lambda$ of $A$.
Size means absolute value, so $-4$ beats $3$. Here's what the dominant eigenvalue does to the iterates.
Let $A$ be an $n \times n$ matrix with a basis of eigenvectors $v_1, \ldots, v_n$, $Av_i = \lambda_iv_i$, where $\lambda_1$ is dominant. If $x_0 = c_1v_1 + \cdots + c_nv_n$ with $c_1 \neq 0$, then
$$\frac{A^kx_0}{\lambda_1^k} \to c_1v_1 \quad \text{as } k \to \infty.$$So for large $k$, $A^kx_0 \approx c_1\lambda_1^kv_1$: the iterates line up along $v_1$, and each step multiplies them by about $\lambda_1$. For a recurrence whose polynomial has different roots with $r_1$ dominant and $c_1 \neq 0$, $a_{k+1}/a_k \to r_1$.
By the powers theorem, $A^kx_0/\lambda_1^k = c_1v_1 + \sum_{i \ge 2} c_i(\lambda_i/\lambda_1)^kv_i$. Each $|\lambda_i/\lambda_1| < 1$, so $(\lambda_i/\lambda_1)^k \to 0$, and the sum goes to $0$. For the recurrence, $a_k = r_1^k(c_1 + \varepsilon_k)$ with $\varepsilon_k = \sum_{i \ge 2} c_i(r_i/r_1)^k \to 0$. Once $|\varepsilon_k| < |c_1|$, $a_k \neq 0$ and $a_{k+1}/a_k = r_1(c_1 + \varepsilon_{k+1})/(c_1 + \varepsilon_k) \to r_1$. $\blacksquare$
Let $A$ be a $2 \times 2$ or $3 \times 3$ matrix with eigenvectors as in the theorem above, where $\lambda_1$ is dominant. If $x_0$ uses $v_1$ (that is, $c_1 \neq 0$), then for large $k$, $A^kx_0 \approx c_1\lambda_1^kv_1$: the iterates line up along $v_1$, and each step multiplies them by about $\lambda_1$. For a recurrence, $a_{k+1}/a_k$ gets close to the root of biggest size.
This is a visual argument, not a formal proof. Shrink every iterate by $\lambda_1^k$, so the winner's leg holds still. Then every other leg gets multiplied by $\lambda_i/\lambda_1$ each step, a number smaller than $1$ in size, so those legs shrink away and the tip slides onto the line of $v_1$. Here it is for the opening figure's $A = \begin{pmatrix} 1 & \frac12 \\ \frac12 & 1 \end{pmatrix}$, where $\lambda_1 = 1.5$ and $\lambda_2 = 0.5$.
The full proof is in the n version.
So a ratio of terms points straight at the dominant root.
Where do the ratios $a_{k+1}/a_k$ go for the running example ($a_0 = 1$, $a_1 = 5$), and for the Fibonacci numbers?
The running example's roots are $2$ and $-1$, and the start uses both ($c_1 = 2$), so the ratios go to $2$: they run $5, \ 1.4, \ 2.43, \ 1.82, \ 2.10, \ 1.95, \ldots$, zigzagging in because the $(-1)^k$ part flips each time.
The Fibonacci roots are $\varphi \approx 1.618$ and $\overline{\varphi} \approx -0.618$, so $F_{k+1}/F_k \to \varphi$: $\frac{13}{8} = 1.625$, $\frac{21}{13} \approx 1.615$, $\frac{34}{21} \approx 1.619$. The ratio of neighboring Fibonacci numbers is the golden ratio, in the limit.
The theorem has two hypotheses, and each one can fail.
Two traps. If the start skips the dominant eigenvector ($c_1 = 0$), the next eigenvalue in line takes over: $a_{k+2} = a_{k+1} + 2a_k$ with $a_0 = 1$, $a_1 = -1$ is just $(-1)^k$, and the $2$ never shows up. And if two eigenvalues tie in size, there may be no limit at all: $a_{k+2} = 4a_k$ (roots $2$ and $-2$) with $a_0 = 0$, $a_1 = 1$ runs $0, 1, 0, 4, 0, 16, \ldots$, and half its ratios can't even be computed.
One more case is the subject of its own page: eigenvalues that aren't real at all, like the rotation's $\pm i$. Then the iterates turn instead of lining up; that's the story of Complex Eigenvalues.
Try it. Drag the start and watch where the dots go.
On an eigenvector's line the dots never leave it; anywhere else they bend toward the line that grows fastest.
Run this on a computer, rescale the arrow after every step so the numbers don't overflow, and its direction settles onto the dominant eigenvector. That's the power method, a later page, and it's how computers find the biggest eigenvalue of enormous matrices.
Your turn: find the winner. In free response, type one number (for a percent, just the number, like 20).
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Not every recurrence is that tidy. In the Tower of Hanoi puzzle you move a stack of disks from one peg to another, one disk at a time, never putting a bigger disk on a smaller one. Moving $k + 1$ disks means moving the top $k$ out of the way, moving the big one, and moving the $k$ back on top of it. So if $a_k$ is the number of moves for $k$ disks, $a_{k+1} = 2a_k + 1$, with $a_0 = 0$. That $+1$ isn't linear: no matrix sends $a$ to $2a + 1$.
Slides had the same problem: a slide of the plane isn't linear, so Computer Graphics lifted the plane to height $1$ and the slide became a shear. Lift the sequence the same way.
Write $a_{k+1} = 2a_k + 1$, $a_0 = 0$, as $x_{k+1} = Ax_k$, and solve it.
Take $x_k = (a_k, 1)$. Then $x_{k+1} = (2a_k + 1, 1)$, so $A = \begin{pmatrix} 2 & 1 \\ 0 & 1 \end{pmatrix}$: the top row does the work and the bottom row keeps the $1$ a $1$. $A$ is triangular, so its eigenvalues are $2$ and $1$, with eigenvectors $(1, 0)$ and $(-1, 1)$.
Split the start: $x_0 = (0, 1) = (1, 0) + (-1, 1)$. So $x_k = 2^k(1, 0) + 1^k(-1, 1) = (2^k - 1, 1)$, and $a_k = 2^k - 1$: $0, 1, 3, 7, 15, \ldots$ The constant $+1$ turned into an eigenvalue $1$, the part that never changes.
Legend says the priests in the original story have $64$ disks. That's $2^{64} - 1 \approx 1.8 \times 10^{19}$ moves: at one move a second, about $585$ billion years.
Ecologists count a population by age class: how many are young, how many are older, year after year. Births and survival are rates, so next year's counts are linear in this year's: $x_{k+1} = Ax_k$.
A Leslie matrix for $m$ age classes has the birth rates $f_1, \ldots, f_m \ge 0$ across its first row, the survival rates $s_1, \ldots, s_{m-1} > 0$ just below the diagonal, and $0$s everywhere else. If $x_k$ counts each age class in year $k$, then $x_{k+1} = Ax_k$.
Births go across the top row, because every class can have babies and the babies all land in the first class. Survival sits just below the diagonal, because surviving moves you up one age. It's the recurrence matrix's shape again, with rates in place of the $1$s.
A bird species has young birds and old birds. Each young bird raises $1$ chick a year on average and each old bird $1.5$; half of the young live to be old, and old birds don't survive the year. So $A = \begin{pmatrix} 1 & 1.5 \\ 0.5 & 0 \end{pmatrix}$. Starting from $20$ young and $20$ old, what happens in the long run?
$\det(A - tI) = t^2 - t - 0.75 = (t - 1.5)(t + 0.5)$, so the eigenvalues are $1.5$ (dominant) and $-0.5$. For $1.5$, the second row of $Av = 1.5v$ says $0.5y = 1.5o$, so $y = 3o$ and $v_1 = (3, 1)$. For $-0.5$, $v_2 = (-1, 1)$.
The start splits as $(20, 20) = 10(3, 1) + 10(-1, 1)$, so
$$x_k = 10 \cdot 1.5^k\begin{pmatrix} 3 \\ 1 \end{pmatrix} + 10(-0.5)^k\begin{pmatrix} -1 \\ 1 \end{pmatrix}.$$The second part dies away. In the long run the population grows by $50\%$ a year, and it settles into $3$ young birds for every old one: $75\%$ young, $25\%$ old.
Watch the mix settle while the total keeps growing.
The counts never settle down, but the shape of the population does, and then it grows at a steady rate (Hefferon calls this a dynamic stability). That shape gets a name.
When a Leslie matrix has a dominant eigenvalue $\lambda_1 > 0$, its stable age distribution is the eigenvector for $\lambda_1$ scaled so its entries add up to $1$: the share of the population in each age class in the long run.
For two classes the eigenvector is quick: the second row of $Av = \lambda v$ says $s \cdot y = \lambda \cdot o$, so $v = (\lambda, s)$. Your turn. In free response, type a number; fractions like 3/4 are fine.
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Three age classes work the same way, one row at a time.
A fish population has newborns, one-year-olds, and two-year-olds, with $A = \begin{pmatrix} 0 & 3 & 4.5 \\ 0.5 & 0 & 0 \\ 0 & 0.5 & 0 \end{pmatrix}$. Show that $1.5$ is the dominant eigenvalue and find the stable age distribution.
Expanding along the first row, $\det(A - tI) = -t^3 + 1.5t + 1.125$, and
$$t^3 - 1.5t - 1.125 = (t - 1.5)(t^2 + 1.5t + 0.75).$$The quadratic has no real roots; its two complex roots have size $\sqrt{0.75} \approx 0.87$, less than $1.5$. So $1.5$ is dominant. For the eigenvector, rows $2$ and $3$ of $Av = 1.5v$ say $0.5v_1 = 1.5v_2$ and $0.5v_2 = 1.5v_3$: each class is a third of the one before. So $v = (9, 3, 1)$, which adds up to $13$, and the stable age distribution is $\left(\frac{9}{13}, \frac{3}{13}, \frac{1}{13}\right)$: about $69\%$ newborns, $23\%$ one-year-olds, $8\%$ two-year-olds, growing $50\%$ a year.
In space, each year's population is an arrow. Drawn at one length, its direction wobbles (that's the complex pair) and then settles onto the line through $(9, 3, 1)$.
Not every Leslie matrix has a winner, though.
$A = \begin{pmatrix} 0 & 4 \\ 0.25 & 0 \end{pmatrix}$ has eigenvalues $1$ and $-1$, the same size, so its counts swap between two patterns forever and never settle. When two neighboring age classes both have babies, that can't happen (a theorem beyond this page), and there is a dominant eigenvalue.
Now run a population yourself. Chicks don't breed here; adults raise $f$ chicks each, a share $s$ of chicks grow up, and a share $a$ of adults survive another year.
The toolkit for this page: a linear recurrence becomes $x_{k+1} = Ax_k$ with the coefficients across the top row, and its eigenvalues are the roots of its polynomial. Split $x_0$ along eigenvectors and $A^kx_0 = \sum c_i\lambda_i^kv_i$, which for a recurrence means $a_k = \sum c_ir_i^k$. In the long run the dominant eigenvalue wins: the iterates line up with its eigenvector and grow by about $\lambda_1$ a step, as long as the start uses it.
Every question so far came right after its idea, which is a hint a quiz never gives you. Down here the hint is gone. First, true or false: a claim from anywhere on this page, and you decide whether it holds.
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Next, a mixed set: every question type on this page, shuffled. Type a number, a matrix, or a formula when it asks you to.
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Made it this far? Time for something with a little heat: a recurrence of order $3$, and a term far down the line. In free response, type one number.
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