Open Problem: Application of Cayley-Hamilton Theorem

DISCLAIMER: This problem is explained properly on my YouTube channel, at the following link: PROBLEM: Analysis of 2x2 Matrices using Cayley-Hamilton Theorem. I suggest you watch this before looking below.

Before I start, let's recall the Cayley-Hamilton Theorem:

Theorem (Cayley-Hamilton Theorem) If AA is a square matrix and cA(λ)=det(λIA)c_A(\lambda) = \det(\lambda I-A) is its characteristc polynomial, then (allowing for matrices as the polynomial variable) cA(A)=0c_A(A) = \bf{0}, where 0\bf{0} is the zero square matrix.

Suppose now that AA is the following 2 x 2 matrix:

A=(abcd) A = \begin{pmatrix} a & b\\ c & d \end{pmatrix}

Note that the trace (denoted by tt) and determinant (denoted by Δ\Delta) is t=a+d t = a + d and Δ=adbc \Delta = ad-bc respectively. Then the characteristic polynomial of AA is:

cA(λ)=det(λIA)=det(λabcλd)=(λa)(λd)bc=λ2tλ+Δ \begin{aligned} c_{A}(\lambda) &= \det(\lambda I - A) = \det \begin{pmatrix} \lambda - a & -b \\ -c & \lambda - d \end{pmatrix} \\ &= (\lambda - a)(\lambda - d) - bc \\ &= \lambda^2 - t \lambda + \Delta \end{aligned}

Thus, by the Cayley-Hamilton Theorem:

A2tA+Δ=0 A^2 - tA + \Delta = \bf{0}


A2=tAΔ A^2 = tA - \Delta

What we can see is that A2A^2 can be expressed in terms of AA, tt and Δ\Delta. Can we express A3A^3 in terms of AA, tt and Δ\Delta? Well, let's compute this:

A3=AA2=A(tAΔ)=tA2ΔA=t(tAΔ)ΔA=(t2Δ)AtΔ \begin{aligned} A^3 &= AA^2 = A(tA - \Delta) \\ &= tA^2 - \Delta A = t(tA - \Delta) - \Delta A \\ &= (t^2 - \Delta)A - t\Delta \end{aligned}

So here's two open problems for you:

1) How to express AnA^n in terms of AA, tt and Δ\Delta?

2) Can we do something similar for square matrix of arbitrary sizes? How about a 3 x 3 matrix?

Note by A Former Brilliant Member
2 years, 11 months ago

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For problem 1, here's an outline of an idea that I think might work, though I'm not very sure. Also, I think this method will give the closed form in the form of some cases with conditions imposed on tt and Δ\Delta.

Denote xn=Anx_n=A^n and define A0=IA^0=I where II is the identity matrix of the same order as AA. Then we have,

xn=An=A2An2=(tAΔ)An2=tAn1ΔAn2=txn1Δxn2  n2(1)x_n=A^n=A^2A^{n-2}=(tA-\Delta)A^{n-2}=tA^{n-1}-\Delta A^{n-2}=tx_{n-1}-\Delta x_{n-2}~\forall~n\geq 2\qquad(1)

Since tt and Δ\Delta only depend on AA which is independent of nn, we note that t,Δt,\Delta are independent of nn and hence (1)(1) can be seen as a linear recurrence relation with constant coefficients and we can use the method described here to get a closed form for xnx_n, i.e., a closed form for AnA^n in terms of t,Δ,At,\Delta,A using the initial values x0=Ix_0=I and x1=Ax_1=A.

I tried working on the case where the auxillary equation for the recurrence has distincts non-zero roots, i.e., when Δ0\Delta\neq 0 and t2Δt^2\neq\Delta, and let me tell you, the result is not very pretty. We have,

An=c1αn+c2βnA^n = c_1\alpha^n + c_2\beta^n

where α,β=t±t24Δ2\alpha,\beta=\dfrac{-t\pm\sqrt{t^2-4\Delta}}2 and c1=1βα(βIA)c_1=\dfrac 1{\beta-\alpha}(\beta I-A) and c2=1βα(αI+A)c_2=\dfrac 1{\beta-\alpha}(-\alpha I+A)

I think there might be a better way to approach the problem but this is all I have right now. Will update the comment if I figure out a better way.

Prasun Biswas - 2 years, 11 months ago

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I guess Problem 1 is almost solved; once you have the linear recurrence, just find the generating function and go from there (I know I'm projecting, and I'm expecting some sort of sh*tstorm!).

Any ideas on Problem 2?

A Former Brilliant Member - 2 years, 11 months ago

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Essentially yes. What you want is the characteristic polynomial of the matrix which arises from cayley hamilton theorem. Say it has degree dn d \leq n .

Then, every matrix Ak A^k can be expressed as I,A,A2,,ad1 I, A, A^2, \ldots, a^{d-1} by repeated substituting in the characteristic polynomial. You could use the linear recurrence relations to solve for each coefficient if desired.

Calvin Lin Staff - 2 years, 11 months ago

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