[HN Gopher] Powers of a 2x2 matrix in closed form
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Powers of a 2x2 matrix in closed form
Author : KqAmJQ7
Score : 54 points
Date : 2023-06-11 10:39 UTC (12 hours ago)
(HTM) web link (www.johndcook.com)
(TXT) w3m dump (www.johndcook.com)
| paulpauper wrote:
| diagonalization, recurrences, system of equations, etc.
|
| there are many ways to do it
| fweimer wrote:
| Any idea what the derivation in [1] looks like?
| edflsafoiewq wrote:
| Let a and b be the eigenvalues of the matrix A. A satisfies its
| characteristic polynomial so we have A^2 - (a +
| b)A + abI = 0
|
| If a /= b, define X = (A - bI)/(a - b) Y
| = (A - aI)/(b - a)
|
| which satisfy X^2 = X XY = YX = 0
| Y^2 = Y
|
| so for k >= 2, X^k = X and Y^k = Y. Hence A^n =
| (aX + bY)^n = a^n X^n + b^n Y^n = a^n X +
| b^n Y
|
| On the other hand, if a = b, define Z = A - aI
|
| which satisfies Z^2 = 0, so Z^k = 0, and A^n =
| (aI + Z)^n = a^n I + n a^(n-1) Z
| fweimer wrote:
| Thanks, that actually looks appropriately elementary.
| [deleted]
| naillo wrote:
| A less formula driven approach that I expected this to use is
| just eigenvalue decompose the matrix (with complex eigenvalues
| that always exists) then raise each eigenvalue to the power you
| want and then combine the eigenvectors and eigenvalues together
| again to get the Nth power. (Requires computing a matrix matrix
| product but only once and that could be expressed in closed form
| and it's overall a cheap operation vs actually multiplying the
| matrix by itself N times.)
| JadeNB wrote:
| > A less formula driven approach that I expected this to use is
| just eigenvalue decompose the matrix (with complex eigenvalues
| that always exists) then raise each eigenvalue to the power you
| want and then combine the eigenvectors and eigenvalues together
| again to get the Nth power.
|
| That's actually what's happening here for the distinct-
| eigenvalues case: (A - \beta I)/(\alpha - \beta) is the
| projection on the \alpha-eigenspace, where A^n acts by
| \alpha^n, with kernel the \beta-eigenspace; and similarly with
| \alpha and \beta switched. In general, pursuing this same line
| of attack underlies the proof that the semisimple part of A can
| be written as a polynomial in A. (Sadly, the polynomial itself
| depends on A, so this is probably more usefully stated as that
| the semisimple part of A lies in k[A] if k is perfect.)
| henrydark wrote:
| It sounds like you're assuming diagonalizability, which is
| stronger than "eigen values defined over the field".
|
| Even over the complex numbers there are matrices that aren't
| diagonalizable, like [[2,1],[0,2]].
| xyzzyz wrote:
| In practice, all matrices are diagonalizable, because the set
| of non-diagonalizable matrices is very small. Concretely, a
| random matrix is almost surely diagonalizable (ie. with
| probability 1).
| FooBarBizBazz wrote:
| This "with zero probability" statement hides that there's a
| probability distribution assumed. It's really saying that
| any matrix can be perturbed to be diagonalizable. Which is
| true. But matrices in that set have a way of showing up a
| lot in models that the human mind conceives. For example,
| if you model a point mass accelerating without drag, you
| will end up with a matrix that isn't diagonalizable. If all
| matrices were diagonalizable, every linear ODE would be
| solved by a bunch of exponentials and sinusoids; there'd be
| no polynomials. But we see things like "1/2 a t^2" all the
| time.
| henrydark wrote:
| Also "any matrix can be perturbed" assumes a normed
| field, though those equations work over other fields as
| well
| [deleted]
| JadeNB wrote:
| > It sounds like you're assuming diagonalizability, which is
| stronger than "eigen values defined over the field".
|
| > Even over the complex numbers there are matrices that
| aren't diagonalizable, like [[2,1],[0,2]].
|
| While that's completely true, and I'm glad someone said it,
| it's not much of an issue for 2x2 matrices. You can tell
| quickly when it's happened, because the characteristic
| polynomial is a square (itself easily tested by computing the
| gcd with the derivative ... or just because it's easy to
| recognize when a quadratic is a perfect square) but the
| matrix isn't scalar; and then we have that the semisimple
| part S is scalar, the nilpotent part N satisfies N^2 = 0, and
| so (S + N)^n equals S^n + n S N^{n - 1} for n > 1.
| FooBarBizBazz wrote:
| Thank you! This why the Jordan Form exists at all.
|
| (All _real symmetric_ , or more generally _Hermitian_
| matrices are diagonalizable, which is the Spectral Theorem.)
| tptacek wrote:
| That would be less interesting, since it's the method taught in
| every first course of linear algebra, right?
| JadeNB wrote:
| > That would be less interesting, since it's the method
| taught in every first course of linear algebra, right?
|
| But it _would_ be interesting to mention that that 's what
| this method is doing, just using that, in the distinct-
| eigenvalues case, A - \beta I maps onto the \alpha-
| eigenspace, so you can simply apply it instead of explicitly
| finding eigenvectors. In fact, every time I teach linear
| algebra, one of my students discovers this 2x2 trick for
| finding eigenvectors, though usually accidentally thinking
| that they are finding \beta-eigenvectors, rather than \alpha-
| eigenvectors, by computing the image of A - \beta I instead
| of its kernel.
| tptacek wrote:
| Oh. Oh, neat. Thanks!
| yxhuvud wrote:
| I must be missing something as this seems very basic.
|
| Assume A = vEv^T , where E is a diagonal matrix of eigenvalues
| and v is the transformation from A to E. Then A^n =
| vEv^TvEv^T..vEv^T. vv^T equals the identity matrix and will
| cancel out. So A^n = vE^nv^T. The closed form in the article is
| then just inlining the computation of v and v^T to how they apply
| to alpha and beta.
|
| Oh, and to understand what happens in the more complex case where
| not all eigenvalues are different, see Jordan Normal Form. That
| is less basic and not as well covered in the basic courses as the
| above is.
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