Powers of a 2×2 matrix in closed form

Here’s something I found surprising: the powers of a 2×2 matrix have a fairly simple closed form. Also, the derivation is only one page [1].

Let A be a 2×2 matrix with eigenvalues α and β. (3Blue1Brown made a nice jingle for finding the eigenvalues of a 2×2 matrix.)

If α = β then the nth power of A is given by

$A^n = \alpha^{n-1}\left( nA - (n-1)\alpha I\right)$

If α ≠ β then the nth power of A is given by

$A^n = \frac{\alpha^n}{\alpha - \beta} (A - \beta I) + \frac{\beta^n}{\beta-\alpha}(A - \alpha I)$

Example

Let’s do an example with

$A = \begin{bmatrix} 6 & 3 \\ 20 & 23 \end{bmatrix}$

The eigenvalues are 26 and 3. I chose the matrix entries based on today’s date, not to have integer eigenvalues, and was surprised that they turned out so simple [2]. (More along those lines here.)

Here’s a little Python code to show that the formula above gives the same result as directly computing the cube of A.

    import numpy as np
    
    A = np.matrix([[6, 3], [20, 23]])
    m = (6 + 23)/2
    p = 6*23 - 3*20
    α = m + (m**2 - p)**0.5
    β = m - (m**2 - p)**0.5
    print(α, β)
    
    I = np.eye(2)
    direct = A*A*A
    
    formula = α**3*(A - β*I)/(α - β) + β**3*(A - α*I)/(β - α)
    print(direct)
    print(formula)

[1] Kenneth S. Williams. The nth Power of a 2×2 Matrix. Mathematics Magazine, Dec., 1992, Vol. 65, No. 5, p. 336.

[2] I wrote a script to find out how often this happens, and it’s more often than I would have guessed. There are 31 dates this year that would give integer eigenvalues if arranged as in the example.

2 thoughts on “Powers of a 2×2 matrix in closed form”

Michael Scharrer

11 June 2023 at 04:29

Interesting. It looks like the α≠β version splits up the matrix A into two rank-1 matrices, each corresponding to an eigenvector. From then on it’s simple to appropriately scale those rank-1 components. Is there a similar intuitive explanation for the α=β version?

Also, both formulae seem to work with n=-1 to give the inverse of non-singular matrices.

Nick Craig-Wood

11 June 2023 at 10:43

Thank you for an interesting article.

I just realized that you can use this result along with the 2×2 generator matrix for Fibonacci numbers [ 1 1 ] [ 1 0 ] to prove Binet’s formula in a very straightforward way.

The eigenvalues of that matrix are 1/2 (1 +/- sqrt(5)) so Binet’s formula appears quite neatly.

Comments are closed.