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# 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 $$A$$ is a square matrix and $$c_A(\lambda) = \det(\lambda I-A)$$ is its characteristc polynomial, then (allowing for matrices as the polynomial variable) $$c_A(A) = \bf{0}$$, where $$\bf{0}$$ is the zero square matrix.

Suppose now that $$A$$ is the following 2 x 2 matrix:

$$A = \begin{pmatrix} a & b\\ c & d \end{pmatrix}$$

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

$$\begin{split} c_{A}(\lambda) &= \det(\lambda I - A) = \det \begin{pmatrix} \lambda - a & -b \\ -c & \lambda - d \end{pmatrix} \\ &= (\lambda - a)(\lambda - d) - bc \\ &= \lambda^2 - t \lambda + \Delta \end{split}$$

Thus, by the Cayley-Hamilton Theorem:

$$A^2 - tA + \Delta = \bf{0}$$

or:

$$A^2 = tA - \Delta$$

What we can see is that $$A^2$$ can be expressed in terms of $$A$$, $$t$$ and $$\Delta$$. Can we express $$A^3$$ in terms of $$A$$, $$t$$ and $$\Delta$$? Well, let's compute this:

$$\begin{split} A^3 &= AA^2 = A(tA - \Delta) \\ &= tA^2 - \Delta A = t(tA - \Delta) - \Delta A \\ &= (t^2 - \Delta)A - t\Delta \end{split}$$

So here's two open problems for you:

1) How to express $$A^n$$ in terms of $$A$$, $$t$$ and $$\Delta$$?

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

10 months, 3 weeks 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 $$t$$ and $$\Delta$$.

Denote $$x_n=A^n$$ and define $$A^0=I$$ where $$I$$ is the identity matrix of the same order as $$A$$. Then we have,

$x_n=A^n=A^2A^{n-2}=(tA-\Delta)A^{n-2}=tA^{n-1}-\Delta A^{n-2}=tx_{n-1}-\Delta x_{n-2}~\forall~n\geq 2\qquad(1)$

Since $$t$$ and $$\Delta$$ only depend on $$A$$ which is independent of $$n$$, we note that $$t,\Delta$$ are independent of $$n$$ and hence $$(1)$$ 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 $$x_n$$, i.e., a closed form for $$A^n$$ in terms of $$t,\Delta,A$$ using the initial values $$x_0=I$$ and $$x_1=A$$.

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

$A^n = c_1\alpha^n + c_2\beta^n$

where $$\alpha,\beta=\dfrac{-t\pm\sqrt{t^2-4\Delta}}2$$ and $$c_1=\dfrac 1{\beta-\alpha}(\beta I-A)$$ and $$c_2=\dfrac 1{\beta-\alpha}(-\alpha I+A)$$

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. · 10 months, 3 weeks ago

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? · 10 months, 3 weeks ago

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

Then, every matrix $$A^k$$ can be expressed as $$I, A, A^2, \ldots, a^{d-1}$$ by repeated substituting in the characteristic polynomial. You could use the linear recurrence relations to solve for each coefficient if desired. Staff · 10 months, 2 weeks ago