Cayley-Hamilton Theorem

Theorem

Theorem (Cayley-Hamilton, 1). For a finite matrix J, with eigenvalues m_1, m_2, ..., m_i, ..., we have the following equation:

(J-m_1 I)(J-m_2 I)...(J-m_i I)... = O

or the following equivalent (and more standard) form:

Theorem (Cayley-Hamilton, 2). For a finite matrix J, with its eigen-polynomial:

\det (\lambda I-J) = \lambda^n + a_{n-1}\lambda^{n-1} + ... + a_0 = 0

Then we have the same relation with J itself:

J^n + a_{n-1}J^{n-1} + ... + a_0 I = O

Proof

Come on, I’m just a student studying physics, not an expert in abstract algebra! Nor do I care!

Usage

I came across this theorem when studying group theory, or more specifically, the irreducible representation j=1 of the group SO(3), in which we have to calculate something like \exp(-i\beta J_2), with

J_2 = \frac{1}{\sqrt{2}i}\left( \begin{matrix} & 1 & \\ -1 & & 1 \\ & -1 & \end{matrix} \right)

The standard way to do it would be to “Taylor expand” it using “exponential” function, but it would produce infinite terms of J_2^n.

Luckily we already know that J_2 has eigenvalues m = -1, 0, 1; so we have

(J_2 - 1)J_2(J_2 + 1) = O

which gives us:

J_2^3 = J_2

This gives us a way to truncate the whole exponential function: we can now simply assume that

\exp(-i\beta J_2) = a(\beta)I + b(\beta) J_2 + c(\beta) J_2^2

Then we can produce 3 eigenvalue equations: (L.H.S. = R.H.S.)

\begin{cases} \begin{aligned} e^{-i\beta \times (-1)} &= a - b + c \\ e^{-i\beta \times 0} &= a \\ e^{-i\beta \times 1} &= a + b + c \end{aligned} \end{cases}

with which we can deduce a, b, c and express \exp(-i\beta J_2) in finite terms of I, J_2, J_2^2.

More generally, it’s possible to express the highest order term with a linear combination of lower order terms, which enable us to truncate the infinite expansion terms into finite terms (which still can be troublesome and boring, but at least it’s a huge progress).