Let A be an n×n matrix of real elements. The determinantal equation defining its eigenvalues is

det(A−λI) = 0

where I is the n×n identity matrix. This equation in terms of a determinant generates a polynomimal equation p(λ)=0 where p(λ) is called the characteristic polynomial of the matrix. The Cayley-Hamilton Theorem is that if A is substituted for λ in the characteristic polynomial the result is a matrix of zeroes.

Illustration of the theorem

Let A be the matrix

1	2
3	4

Then the matrix A−λI is

1−λ	2
3	4−λ

The determinant of (A−λI) is then

(1−λ)(4−λ)−2*3
which reduces to
4 −5λ+λ² −6
and further to
−2 −5λ + λ²

This is the characteristic polynomial of the matrix A. The solutions to the polynomical equation

−2 −5λ + λ² = 0

are the eigenvalues of the matrix A.

Consider now what results when λ is replaced with A and −2 is replaced with −2I.

A² is the matrix

7	10
15	22

Therefore −2I −5A + A² is

−2−5*1+7	0−5*2+10
0−5*3+15	−2−5*4+22

which evaluates to

0	0
0	0

Thus, almost mystically, A satisfies its own eigenvalue value equation.

Quick Proof

The charactistic polynomial equation is equivalent to the determinantal equation

det(A−λI) = 0

det(A−Λ) = 0

If A is substituted for Λ in this equation the result is

det(A−A) = det(O) = 0

where O is a matrix of zeroes. Certainly A satisfies the determinantal equation.

The Long Tedious Proof

Let the eigenvalue equation be expressed in the equivalent form

det(λI−A) = 0

This removes the ambiguity of the sign of λⁿ in the characteristic equation; it is always 1 for all n.

(To be continued.)