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Powers through the eigenbasis

Raise a matrix to any power by raising its eigenvalues instead: derive A^k = P D^k P inverse, follow a vector’s eigen-coordinates as they grow or shrink, read the long-run behaviour off the eigenvalues, and compare the cost with repeated multiplication.
M03.2.2L2 Intermediate25 min
In review

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Before this lesson: M03.2.1 A = P D P inverse

What you will be able to do
  • Derive A^k = P D^k P^(-1) by cancelling the inner pairs P^(-1) P, and the equivalent form A^k x = c_1 lambda_1^k v_1 + c_2 lambda_2^k v_2.
  • Compute a power of a 2 by 2 matrix by hand both ways and check the two answers agree.
  • Predict from the eigenvalues alone whether the powers of a matrix decay, grow or settle on a steady direction.
  • Apply a polynomial to a diagonalisable matrix through its eigenvalues, f(A) = P f(D) P^(-1).
  • Compare the eigen route with numpy.linalg.matrix_power in operations and in exactness.

The question

A recurrent network applies the same weight matrix at every time step, a Markov chain applies the same transition matrix at every click, and PageRank multiplies by one matrix again and again. What does a matrix do after fifty applications, or a thousand? Multiplying it out is slow and hides the answer. The eigenvalues show it at a glance.

From the last lesson

1. Why does \(\mat{A}\mat{P} = \mat{P}\mat{D}\) hold?

Show answer

Column \(j\) of \(\mat{A}\mat{P}\) is \(\mat{A}\vect{v}_j = \lambda_j\vect{v}_j\), and column \(j\) of \(\mat{P}\mat{D}\) is \(\mat{P}\) times \(\lambda_j\vect{e}_j\), which is also \(\lambda_j\vect{v}_j\).

2. What are \(\mat{P}\), \(\mat{D}\) and \(\mat{P}^{-1}\) for the running matrix \(\mat{A}\)?

Show answer

\(\mat{P} = \begin{bmatrix} 1 & 1 \\ 1 & -2 \end{bmatrix}\), \(\mat{D} = \begin{bmatrix} 5 & 0 \\ 0 & 2 \end{bmatrix}\) and \(\mat{P}^{-1} = \tfrac13\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}\).

Intuition

Write a vector in eigen-coordinates, \(\vect{x} = {c_1\vect{v}_1 + c_2\vect{v}_2}\). One application of \(\mat{A}\) multiplies \(c_1\) by \(\lambda_1\) and \(c_2\) by \(\lambda_2\), and the two never mix. So \(k\) applications multiply them by \(\lambda_1^k\) and \(\lambda_2^k\). A matrix power, seen in the eigenbasis, is two ordinary powers of numbers.

The powers of \(\mat{A}\) itself grow too fast for a picture, so take \(\mat{M} = \tfrac15\mat{A} = \begin{bmatrix} 0.8 & 0.2 \\ 0.4 & 0.6 \end{bmatrix}\). Dividing by 5 keeps the eigenvectors and divides the eigenvalues by 5: they become 1 and 0.4. (Each row of \(\mat{M}\) adds up to 1. Matrices like it are the Markov matrices of module M03.9.)

Two panels. The left panel, titled x k equals M to the k times x 0 in the plane, shows pale orange eigen-lines, a blue arrow to x 0 equals (2, minus 1) and green dots for x 1, x 2 and later iterates, joined by green arrows. They march in a straight dashed line towards the orange dot at v1 equals (1, 1), each step covering less ground than the one before. The right panel, titled Its two eigen-coordinates, plots against the step k from 0 to 5 a flat orange line at height 1, labelled c1 lambda1 to the k equals 1 times 1 to the k: kept, and a dashed orange curve falling from 1 through 0.4 and 0.16 towards 0, labelled c2 lambda2 to the k equals 0.4 to the k, shrinks by 0.4 a step.
M03.2.2-F1

The powers of \(\mat{M} = \frac15\mat{A}\) applied to \(\vect{x}_0 = \vect{v}_1 + \vect{v}_2\). The part along \(\vect{v}_1\) is multiplied by 1 at every step and stays. The part along \(\vect{v}_2\) is multiplied by 0.4 and fades, so the iterates slide along a straight line onto \(\vect{v}_1\).

Every question about the long run of \(\mat{M}^k\vect{x}\) becomes a question about \(1^k\) and \(0.4^k\). The first stays 1; the second goes to 0. So the iterates settle on \(\vect{v}_1 = (1, 1)\), and the picture shows it.

Build it up

Step 1. Square the factorisation. From lesson M03.2.1, \(\mat{A} = \mat{P}\mat{D}\mat{P}^{-1}\). Then

\[ \mat{A}^2 = \mat{P}\mat{D}\,\mat{P}^{-1}\mat{P}\,\mat{D}\mat{P}^{-1}. \]

Step 2. Cancel the inner pair. Matrix products may be regrouped (associativity, M02.5), so group the middle as \(\mat{P}^{-1}\mat{P} = \mat{I}\):

\[ \mat{A}^2 = \mat{P}\mat{D}\,\mat{I}\,\mat{D}\mat{P}^{-1} = \mat{P}\mat{D}^2\mat{P}^{-1}. \]

Step 3. Every power works the same way. In \(\mat{A}^k\) there are \(k\) copies of \(\mat{P}\mat{D}\mat{P}^{-1}\) and \(k - 1\) inner pairs, and each pair cancels.

Three rows. Top: A cubed equals three bracketed copies of P D P inverse side by side, with each P inverse at the end of one copy and the P at the start of the next copy shaded yellow together, and under each shaded pair the note P inverse P equals I. Middle: equals P, D D D, P inverse, which equals P D cubed P inverse, with the note only the middle is raised to the power. Bottom: D cubed equals the diagonal matrix with entries 5 cubed and 2 cubed, which equals the diagonal matrix with entries 125 and 8, with the note no mixing between the two diagonal entries.
M03.2.2-F2

In \(\mat{A}^3\) the two inner pairs \(\mat{P}^{-1}\mat{P}\) cancel, the three copies of \(\mat{D}\) meet in the middle, and a power of a diagonal matrix is the power of each entry.

Step 4. Powers of a diagonal matrix are free. The product of two diagonal matrices is diagonal, with entries multiplied in pairs, so

\[ \mat{D}^k = \begin{bmatrix} \lambda_1^k & 0 \\ 0 & \lambda_2^k \end{bmatrix}. \]

Step 5. The same fact, for one vector. Apply \(\mat{A}\) to \(\vect{x} = {c_1\vect{v}_1 + c_2\vect{v}_2}\). By linearity and \(\mat{A}\vect{v}_j = \lambda_j\vect{v}_j\), \({\mat{A}\vect{x}} = {c_1\lambda_1\vect{v}_1 + c_2\lambda_2\vect{v}_2}\). Apply it again and each term picks up one more factor. After \(k\) steps

\[ \mat{A}^k\vect{x} = c_1\lambda_1^k\,\vect{v}_1 + c_2\lambda_2^k\,\vect{v}_2 . \]

The coefficients \(c_1, c_2\) are the eigen-coordinates of \(\vect{x}\), that is, \((c_1, c_2) = \mat{P}^{-1}\vect{x}\). So this is \(\mat{P}\mat{D}^k\mat{P}^{-1}\vect{x}\) read from the right: find the coefficients, raise the eigenvalues, rebuild.

The formal statement

NoteResult: powers of a diagonalisable matrix

If \(\mat{A} = \mat{P}\mat{D}\mat{P}^{-1}\) with \(\mat{D}\) diagonal, then for every whole number \(k \ge 1\)

\[ \mat{A}^k = \mat{P}\mat{D}^k\mat{P}^{-1}, \]

and for \(\vect{x} = c_1\vect{v}_1 + \dots + c_n\vect{v}_n\),

\[ \mat{A}^k\vect{x} = c_1\lambda_1^k\vect{v}_1 + \dots + c_n\lambda_n^k\vect{v}_n . \]

Proof by induction. For \(k = 1\) the first formula is the factorisation itself. Suppose it holds for some \(k\). Then

\[ \begin{aligned} \mat{A}^{k+1} &= \mat{A}^k\mat{A} \\ &= \mat{P}\mat{D}^k\mat{P}^{-1}\,\mat{P}\mat{D}\mat{P}^{-1} \\ &= \mat{P}\mat{D}^k\mat{D}\mat{P}^{-1} = \mat{P}\mat{D}^{k+1}\mat{P}^{-1}, \end{aligned} \]

so it holds for \(k + 1\). For the second formula, \(\mat{A}^k\vect{v}_j = \lambda_j^k\vect{v}_j\) by the same induction (one more application multiplies by one more \(\lambda_j\)), and \(\mat{A}^k\) is linear, so it acts on each term of the sum separately. \(\blacksquare\)

The long run, read off the eigenvalues. Every term \(c_j\lambda_j^k\vect{v}_j\) behaves like the number \(\lambda_j^k\).

  • If every \(\abs{\lambda_j} < 1\), every term goes to \(\vect{0}\), so \(\mat{A}^k \to \mat{0}\): decay.
  • If some \(\abs{\lambda_j} > 1\) and the start has \(c_j \neq 0\), that term grows without bound: growth, along \(\vect{v}_j\).
  • If the largest eigenvalue is exactly 1 and all others are smaller in size, the \(\lambda = 1\) term stays and the rest fade: a steady state \(c_1\vect{v}_1\).
Three small panels, each with the same pale orange eigen-lines through the origin and the same blue starting point x 0 equals (0, 2), followed by six green iterates joined by green arrows. Left, titled Decay: 0.5 and 0.2: the iterates fall straight into the origin, noted arrow to 0. Middle, titled Steady: 1 and 0.4: the iterates slide onto the eigen-line of v1 and stop at a point on it, noted arrow to two thirds v1. Right, titled Growth: 1.25 and 0.5: the iterates first dip, then run outwards along the eigen-line of v1, noted grows along v1.
M03.2.2-F5

The same start under \(\frac{1}{10}\mat{A}\), \(\frac15\mat{A}\) and \(\frac14\mat{A}\), which share the eigenvectors of \(\mat{A}\) and have eigenvalues 0.5 and 0.2, 1 and 0.4, and 1.25 and 0.5. The largest eigenvalue decides the fate; the smaller one only decides how fast the iterates forget the \(\vect{v}_2\) direction.

A negative or complex eigenvalue of size 1 does not settle: \((-1)^k\) flips forever and a complex number of modulus 1 turns forever. Module M03.9 meets both.

Functions of a matrix. Since \(\mat{A}^j = \mat{P}\mat{D}^j\mat{P}^{-1}\) for each power, and \(\mat{I} = \mat{P}\mat{P}^{-1}\), any polynomial \(f(t) = {a_0 + a_1t + \dots + a_mt^m}\) gives

\[ \begin{aligned} f(\mat{A}) &= a_0\mat{I} + a_1\mat{A} + \dots + a_m\mat{A}^m \\ &= \mat{P}\,(a_0\mat{I} + \dots + a_m\mat{D}^m)\,\mat{P}^{-1} \\ &= \mat{P}\,f(\mat{D})\,\mat{P}^{-1}, \end{aligned} \]

where \(f(\mat{D})\) is diagonal with entries \(f(\lambda_j)\). A striking case: the characteristic polynomial \(p(t) = {t^2 - 7t + 10}\) of \(\mat{A}\) has \(p(5) = p(2) = 0\), so \(p(\mat{D}) = \mat{0}\) and therefore \({\mat{A}^2 - 7\mat{A} + 10\mat{I}} = \mat{0}\) (exercise 3). Replacing the polynomial by the exponential series gives the matrix exponential \(e^{\mat{A}}\), which solves systems of differential equations; module M12.1 takes it up.

Worked example

Compute \(\mat{A}^5\) for \(\mat{A} = \begin{bmatrix} 4 & 1 \\ 2 & 3 \end{bmatrix}\), once with the whole matrix and once through one vector.

Route 1, the matrix. \(\mat{D}^5\) has entries \(5^5 = 3125\) and \(2^5 = 32\). Multiplying by \(\mat{P}\) on the left scales its columns:

\[ \mat{P}\mat{D}^5 = \begin{bmatrix} 3125 & 32 \\ 3125 & -64 \end{bmatrix}. \]

Then multiply by \(\mat{P}^{-1} = \tfrac13\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix}\), row by column:

\[ \begin{aligned} \mat{A}^5 &= \frac13\begin{bmatrix} 6250 + 32 & 3125 - 32 \\ 6250 - 64 & 3125 + 64 \end{bmatrix} \\[4pt] &= \frac13\begin{bmatrix} 6282 & 3093 \\ 6186 & 3189 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 2094 & 1031 \\ 2062 & 1063 \end{bmatrix}. \end{aligned} \]

Two checks from lesson M03.1.3: the trace is \({2094 + 1063} = 3157 = {3125 + 32}\), the sum of the eigenvalues of \(\mat{A}^5\), and the determinant is \({2094 \cdot 1063} - {1031 \cdot 2062} = 100000 = 10^5\).

Route 2, one vector. Take \(\vect{x} = (1, 0)\). Its eigen-coordinates are \({\mat{P}^{-1}(1, 0)} = {(\tfrac23, \tfrac13)}\), so \(\vect{x} = {\tfrac23\vect{v}_1 + \tfrac13\vect{v}_2}\).

Two routes to the fifth power. Route 1, the whole matrix: the orange diagonal matrix D to the fifth with entries 3125 and 32, a green arrow labelled P times leading to the vermillion matrix with rows 3125, 32 and 3125, minus 64, then a green arrow labelled times P inverse leading to a yellow-shaded matrix with rows 2094, 1031 and 2062, 1063, equals A to the fifth, with the note one third of 2 times 3125 plus 32 equals 2094, and so on. Route 2, one vector x equals (1, 0): x equals two thirds v1 plus one third v2, with P inverse x equals (two thirds, one third), a green arrow to two thirds times 5 to the fifth v1 plus one third times 2 to the fifth v2, each part times its own lambda to the fifth, and a green arrow to the yellow-shaded column (2094, 2062), the first column of A to the fifth. A line underneath: one third of (6250 plus 32, 6250 minus 64) equals (2094, 2062): the two routes agree.
M03.2.2-F3

The same power two ways. The whole matrix needs \(\mat{D}^5\) and two products. One vector needs only its two eigen-coordinates, each multiplied by its own eigenvalue to the fifth.

Each coordinate is multiplied by its own eigenvalue five times:

\[ \begin{aligned} \mat{A}^5\vect{x} &= \tfrac23\cdot 3125\,(1, 1) \\ &\quad + \tfrac13\cdot 32\,(1, -2) \\ &= \tfrac13\,(6250 + 32,\; 6250 - 64) \\ &= (2094,\; 2062). \end{aligned} \]

That is the first column of \(\mat{A}^5\), as it must be, because \(\mat{A}^5(1, 0)\) picks out the first column. The two routes agree.

Code

From scratch: the eigen route with exact fractions, against plain repeated multiplication.

from fractions import Fraction

def matmul(X, Y):
    return [[sum(X[i][t] * Y[t][j] for t in range(2)) for j in range(2)] for i in range(2)]

A = [[4, 1], [2, 3]]
P = [[1, 1], [1, -2]]
P_inv = [[Fraction(2, 3), Fraction(1, 3)], [Fraction(1, 3), Fraction(-1, 3)]]

def power_eigen(k):
    Dk = [[5 ** k, 0], [0, 2 ** k]]
    return [[int(e) for e in row] for row in matmul(matmul(P, Dk), P_inv)]

def power_slow(k):
    out = [[1, 0], [0, 1]]
    for _ in range(k):
        out = matmul(out, A)
    return out

print("A^5, eigen route:", power_eigen(5))
print("A^5, slow route: ", power_slow(5))
A^5, eigen route: [[2094, 1031], [2062, 1063]]
A^5, slow route:  [[2094, 1031], [2062, 1063]]

How much work is each route? The slow route uses \(k - 1\) matrix products. NumPy’s matrix_power uses repeated squaring: write \(k\) in binary, square the matrix once per binary digit, and multiply in the squares whose digit is 1. The eigen route, once \(\mat{P}\) and \(\mat{P}^{-1}\) are known, always needs two matrix products, whatever \(k\) is.

def squaring_count(k):
    """Matrix products used by repeated squaring for the power k: (squarings, extra products)."""
    bits = bin(k)[2:]
    return len(bits) - 1, bits.count("1") - 1

for k in (5, 20, 1000):
    squares, extra = squaring_count(k)
    print(f"k = {k:4d}: slow {k - 1:3d} products, squaring {squares + extra:2d}, eigen route 2")
k =    5: slow   4 products, squaring  3, eigen route 2
k =   20: slow  19 products, squaring  5, eigen route 2
k = 1000: slow 999 products, squaring 14, eigen route 2

For \(k = 1000\): 999 products, 14 products, or 2. Now the library. With an integer array matrix_power is exact, but NumPy’s 64-bit integers hold numbers only up to about \({9.2 \times 10^{18}}\), and the entries of \(\mat{A}^{28}\) are bigger. NumPy does not warn: the answer wraps around and is simply wrong.

import numpy as np

A_np = np.array([[4, 1], [2, 3]])
print("A^5:\n", np.linalg.matrix_power(A_np, 5))

def exact_power(k):
    """Entries of A^k from the eigen formula, in Python's unlimited integers."""
    return [[(2 * 5 ** k + 2 ** k) // 3, (5 ** k - 2 ** k) // 3],
            [(2 * 5 ** k - 2 * 2 ** k) // 3, (5 ** k + 2 * 2 ** k) // 3]]

for k in (27, 28):
    print(k, "int64 correct?", np.linalg.matrix_power(A_np, k).tolist() == exact_power(k))

P_f = np.array([[1.0, 1.0], [1.0, -2.0]])
eigen_30 = P_f @ np.diag([5.0 ** 30, 2.0 ** 30]) @ np.linalg.inv(P_f)
print("float eigen route, A^30 entry (1, 2):", f"{eigen_30[0, 1]:.0f}")
print("exact:                               ", exact_power(30)[0][1])
A^5:
 [[2094 1031]
 [2062 1063]]
27 int64 correct? True
28 int64 correct? False
float eigen route, A^30 entry (1, 2): 310440858204801597440
exact:                                310440858204801591267

The float eigen route is fast but approximate: beyond \(2^{53}\) a float cannot even hold every whole number, so the last digits of \(\mat{A}^{30}\) are lost. Python’s own integers never overflow, which is why the exact formula above uses them.

Explore

M03.2-X1

Power stepper

Choose a matrix, drag the start (or use its sliders) and press step. The bars show the two eigen-coordinates \(c_1\lambda_1^k\) and \(c_2\lambda_2^k\). Switch on normalising to run the power method of lesson M03.2.4.

A static picture in place of the interactive: on the left, the iterates of x 0 equals (2, minus 1) under M equals one fifth of A sliding along a straight line onto the eigenvector v1 equals (1, 1); on the right, its two eigen-coordinates against the step k, one staying at 1 and the other shrinking by 0.4 each step.

Things to try:

  • With the running matrix \(\mat{A}\), step five times from \((1, 0)\) and compare the bars with route 2 of the worked example: \(\tfrac23 \cdot 5^5\) and \(\tfrac13 \cdot 2^5\).
  • Choose \(\tfrac15\mat{A}\) and step until the second bar has all but vanished. The vector stops moving: a steady state.
  • Choose the swap matrix, whose eigenvalues are 1 and \(-1\). The second bar flips sign at every step and never shrinks.

Pitfalls

Two line charts of the length of N to the k times x against the step k from 0 to 10, on a logarithmic scale from 0.3 to 30, for N with rows 0.5, 0.9 and 0.9, 0.5 and x equals (1, 0). The left chart, marked with a cross and titled Judged by the entries, shows the blue dots rising from 1 to about 20 while a dashed reddish-purple line labelled guess: 0.9 to the k falls below 1. The right chart, marked with a tick and titled Judged by the eigenvalues, shows the same blue dots lying on a dashed green straight line labelled approximately 0.71 times 1.4 to the k, with the note lambda equals 1.4 and minus 0.4.
M03.2.2-F4

Wrong beside right. Every entry of \(\mat{N}\) is below 1, yet its powers grow, because its eigenvalues are 1.4 and \(-0.4\). On a log scale the lengths lie on the line \(0.71 \cdot 1.4^k\).

“All the entries are smaller than 1, so the powers shrink.”

Size is decided by the eigenvalues, not by the entries. The matrix \(\mat{N} = \begin{bmatrix} 0.5 & 0.9 \\ 0.9 & 0.5 \end{bmatrix}\) sends \((1, 1)\) to \({(1.4, 1.4)}\), so \(\lambda = 1.4\) and its powers grow like \(1.4^k\). The same mistake in the other direction is just as easy: large entries can come with small eigenvalues. This is why recurrent networks are analysed through the eigenvalues of their weight matrices.

“\(\mat{A}^k = \mat{P}^k\mat{D}^k\mat{P}^{-k}\).”

Only the middle factor is raised to the power. The outer \(\mat{P}\) and \(\mat{P}^{-1}\) appear once each, because all the inner ones cancelled. Raising \(\mat{P}\) to a power would change the basis \(k\) times over, which has nothing to do with \(\mat{A}\).

“The eigen route is exact, so I can use it for big integer powers.”

In floating point it is not: by \(\mat{A}^{30}\) the entries are far beyond \(2^{53}\) and the last digits are lost. And matrix_power on an int64 array is exact only until the entries pass about \({9.2 \times 10^{18}}\), which for \(\mat{A}\) happens at \(k = 28\), without a warning. For exact large powers use Python integers or dtype=object.

Check yourself

Recall

1. Why does \(\mat{A}^k\) equal \(\mat{P}\mat{D}^k\mat{P}^{-1}\)?

Show answer

In the product of \(k\) copies of \(\mat{P}\mat{D}\mat{P}^{-1}\), every inner pair \(\mat{P}^{-1}\mat{P}\) is the identity and drops out, leaving \(\mat{P}\mat{D}^k\mat{P}^{-1}\).

2. If \(\vect{x} = c_1\vect{v}_1 + c_2\vect{v}_2\) in eigenvectors, what is \(\mat{A}^k\vect{x}\)?

Show answer

\(c_1\lambda_1^k\vect{v}_1 + c_2\lambda_2^k\vect{v}_2\): each eigen-coordinate is multiplied by its own eigenvalue \(k\) times.

3. What do the eigenvalues say about the long run of \(\mat{A}^k\vect{x}\)?

Show answer

All of size below 1: decay to zero. One above 1 (with a non-zero coefficient): growth along its eigenvector. Largest exactly 1 and the others smaller: a steady state along the eigenvector of 1.

4. What are the first column of \(\mat{A}^5\) and its trace for the running \(\mat{A}\)?

Show answer

The first column is \((2094, 2062)\), and the trace is \(3157 = 5^5 + 2^5\).

5. How is a polynomial \(f\) applied to a diagonalisable matrix?

Show answer

\(f(\mat{A}) = \mat{P}f(\mat{D})\mat{P}^{-1}\), where \(f(\mat{D})\) is diagonal with entries \(f(\lambda_j)\).

Exercises

1. ★ Compute \(\mat{A}^3\) through the eigenbasis and check it by multiplying \(\mat{A}^2 = \begin{bmatrix} 18 & 7 \\ 14 & 11 \end{bmatrix}\) by \(\mat{A}\).

Show solution

\(\mat{D}^3\) has entries 125 and 8, so \(\mat{P}\mat{D}^3 = \begin{bmatrix} 125 & 8 \\ 125 & -16 \end{bmatrix}\). Then

\[ \begin{aligned} \mat{A}^3 &= \frac13\begin{bmatrix} 250 + 8 & 125 - 8 \\ 250 - 16 & 125 + 16 \end{bmatrix} \\[4pt] &= \begin{bmatrix} 86 & 39 \\ 78 & 47 \end{bmatrix}. \end{aligned} \]

Check: \(\mat{A}^2\mat{A}\) has first row \({(18 \cdot 4 + 7 \cdot 2,\; 18 + 21)} = (86, 39)\) and second row \({(14 \cdot 4 + 11 \cdot 2,\; 14 + 33)} = (78, 47)\).

2. ★★ For \(\mat{M} = \tfrac15\mat{A}\), find the limit of \(\mat{M}^k\) as \(k\) grows, and the limit of \(\mat{M}^k\vect{x}_0\) for \(\vect{x}_0 = (0, 2)\).

Show solution

\(\mat{M}\) has the eigenvectors of \(\mat{A}\) and eigenvalues 1 and 0.4, so \(\mat{M}^k = \mat{P}\begin{bmatrix} 1 & 0 \\ 0 & 0.4^k \end{bmatrix}\mat{P}^{-1}\). As \(0.4^k \to 0\),

\[ \begin{aligned} \mat{M}^k &\to \mat{P}\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\mat{P}^{-1} \\[4pt] &= \begin{bmatrix} 1 & 0 \\ 1 & 0 \end{bmatrix}\frac13\begin{bmatrix} 2 & 1 \\ 1 & -1 \end{bmatrix} \\[4pt] &= \frac13\begin{bmatrix} 2 & 1 \\ 2 & 1 \end{bmatrix}. \end{aligned} \]

So \(\mat{M}^k(0, 2) \to {\tfrac13(0 + 2,\; 0 + 2)} = {(\tfrac23, \tfrac23)} = \tfrac23\vect{v}_1\), the end point of the middle panel of the figure on long runs. Both rows of the limit are \({(\tfrac23, \tfrac13)}\): every start forgets everything except its \(\vect{v}_1\) coordinate.

3. ★★★ Use \(f(\mat{A}) = \mat{P}f(\mat{D})\mat{P}^{-1}\) with \(p(t) = t^2 - 7t + 10\) to show \({\mat{A}^2 - 7\mat{A} + 10\mat{I}} = \mat{0}\). Deduce a formula for \(\mat{A}^{-1}\) that needs no determinant, and check it.

Show solution

\(p(5) = 25 - 35 + 10 = 0\) and \(p(2) = 4 - 14 + 10 = 0\), so \(p(\mat{D}) = \mat{0}\) and \(p(\mat{A}) = \mat{P}\,\mat{0}\,\mat{P}^{-1} = \mat{0}\). Rearranging, \({\mat{A}(7\mat{I} - \mat{A})} = 10\mat{I}\), so

\[ \begin{aligned} \mat{A}^{-1} &= \tfrac{1}{10}(7\mat{I} - \mat{A}) \\ &= \tfrac{1}{10}\begin{bmatrix} 3 & -1 \\ -2 & 4 \end{bmatrix}. \end{aligned} \]

Check: \(\mat{A}\) times \({\begin{bmatrix} 3 & -1 \\ -2 & 4 \end{bmatrix}}\) has rows \({(12 - 2,\; -4 + 4)} = (10, 0)\) and \({(6 - 6,\; -2 + 12)} = (0, 10)\), which is \(10\mat{I}\).

Summary card

In the eigenbasis a matrix power is a power of each eigenvalue.

The anchor figure again: iterates of M equals one fifth of A sliding onto the eigenvector v1, beside the two eigen-coordinates, one staying at 1 and one shrinking by 0.4 a step.
M03.2.2-F1

Each eigen-coordinate on its own.

  • Matrix form. \(\mat{A}^k = \mat{P}\mat{D}^k\mat{P}^{-1}\), with \(\mat{D}^k\) the powers of the eigenvalues.
  • Vector form. \(\mat{A}^k\vect{x} = c_1\lambda_1^k\vect{v}_1 + c_2\lambda_2^k\vect{v}_2\), \((c_1, c_2) = \mat{P}^{-1}\vect{x}\).
  • Long run. All \(\abs{\lambda} < 1\): decay. One \(\abs{\lambda} > 1\): growth. Largest \(= 1\): steady state.
  • Functions. \(f(\mat{A}) = \mat{P}f(\mat{D})\mat{P}^{-1}\) for any polynomial \(f\).
  • Running example. \(\mat{A}^5 = \begin{bmatrix} 2094 & 1031 \\ 2062 & 1063 \end{bmatrix}\).

Sources and next steps

Sources. Each was opened and checked against this lesson on 15 November 2026.

  • Gilbert Strang, Diagonalization and powers of A, MIT OpenCourseWare 18.06SC, lecture 22 and its lecture summary. The summary derives \(A^2 = S\Lambda^2S^{-1}\) by cancelling \(S^{-1}S\) and states \(A^k = S\Lambda^kS^{-1}\), the theorem that \(A^k \to 0\) exactly when every \(\abs{\lambda_i} < 1\) (for \(A\) with \(n\) independent eigenvectors), and the solution \(u_k = c_1\lambda_1^kx_1 + \dots + c_n\lambda_n^kx_n\) of \(u_{k+1} = Au_k\).
  • David Austin, Understanding Linear Algebra, section Diagonalization, similarity, and powers of a matrix. The proposition that \(A = PDP^{-1}\) implies \(A^k = PD^kP^{-1}\), and that \(D^k\) is found by raising each diagonal entry to the power \(k\).
  • Dan Margalit and Joseph Rabinoff, Interactive Linear Algebra, Section 5.4, Diagonalization. Computing \(A^n = CD^nC^{-1}\) once \(A = CDC^{-1}\), with the powers of a diagonal matrix taken entry by entry.
  • NumPy reference, numpy.linalg.matrix_power. For positive integers the power is computed by repeated matrix squarings and matrix multiplications, and the result has the same type as the input for a non-negative exponent.

Next. M03.2.3 Fibonacci and linear recurrences turns a famous sequence into matrix powers and reads off its closed formula.

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M03.2.3 Fibonacci and linear recurrences

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