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Powers of a 2x2 matrix in closed form

Posted on 3 June 2023 by John

Here's something I found surprising: the powers of a 2x2 matrix have
a fairly simple closed form. Also, the derivation is only one page
[1].

Let A be a 2x2 matrix with eigenvalues a and b. (3Blue1Brown made a
nice jingle for finding the eigenvalues of a 2x2 matrix.)

If a = b then the nth power of A is given by

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

If a [?] b 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
    a = m + (m**2 - p)**0.5
    b = m - (m**2 - p)**0.5
    print(a, b)

    I = np.eye(2)
    direct = A*A*A

    formula = a**3*(A - b*I)/(a - b) + b**3*(A - a*I)/(b - a)
    print(direct)
    print(formula)

[1] Kenneth S. Williams. The nth Power of a 2x2 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.

Categories : Math
Tags : Linear algebra
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2 thoughts on "Powers of a 2x2 matrix in closed form"

 1.
    Michael Scharrer
    11 June 2023 at 04:29

    Interesting. It looks like the a[?]b 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 a=b
    version?

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

 2.
    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 2x2
    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.

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