(’1)n ±1 I = ’C1 + AC0 ,

(’1)n ±2 I = ’C2 + AC1 ,

.

.

.

(’1)n ±n’1 I = ’Cn’1 + ACn’2 ,

(’1)n ±n I = ACn’1 .

Multiply the ¬rst equation on the left by An , the second by An’1 , and so

on down the last equation which we multiply by A0 ≡ I. Now add. We ¬nd

that the sum telescopes to give Cayley™s theorem,

An + ±1 An’1 + · · · + ±n I = 0,

as advertised.

A.5.3 Di¬erentiating Determinants

Suppose that the elements of A depend on some parameter x. From the

elementary de¬nition

det A = i1 i2 ...in a1i1 a2i2 . . . anin ,

we ¬nd

d

a1i1 a2i2 . . . anin + a1i1 a2i2 . . . anin + · · · + a1i1 a2i2 . . . anin .

det A = i1 i2 ...in

dx

(A.66)

In other words,

a11 a12 ... a1n

a11 a12 . . . a1n a11 a12 . . . a1n

a21 a22 ... a2n

a21 a22 . . . a2n a21 a22 . . . a2n

d

. +· · ·+

det A = . +. ..

. . . .

.

.. ..

..

. . . . . .

. . .

. .

dx .

. . . . . .

. . .

an1 an2 . . . ann an1 an2 . . . ann

an1 an2 . . . ann

A.6. DIAGONALIZATION AND CANONICAL FORMS 297

The same result can also be written more compactly as

d daij

det A = Aij , (A.67)

dx dx

ij

where Aij is cofactor of aij . Using the connection between the adjugate

matrix and the inverse, this is equivalent to

1d dA ’1

det A = tr , (A.68)

A

det A dx dx

or

d dA ’1

ln (det A) = tr . (A.69)

A

dx dx

A special case of this formula is the result

‚

ln (det A) = A’1 . (A.70)

‚aij ji

A.6 Diagonalization and Canonical Forms

An essential part of the linear algebra tool-kit is the set of techniques for the

reduction of a matrix to its simplest, canonical form. This is often a diagonal

matrix.

A.6.1 Diagonalizing Linear Maps

A common task is the diagonalization of a matrix A representing a linear

map A. Let us recall some standard material relating to this:

i) If Ax = »x, the vector x is said to be an eigenvector of A with eigen-

value ».

ii) A linear operator A on a ¬nite-dimensional vector space is said to be

hermitian, or self-adjoint, with respect to the inner product , if

A = A† , or equivalently x, Ay = Ax, y for all x, y.

iii) If A is hermitian with respect to , , then » is real. To see this,

write

» x, x = x, »x = x, Ax = Ax, x = »x, x = » — x, x . (A.71)

298 APPENDIX A. ELEMENTARY LINEAR ALGEBRA

iii) If A is hermitian and »i and »j are two distinct eigenvalues with eigen-

vectors xi and xj , then xi , xj = 0. To see this, write

»j xi , xj = xi , Axj = Axi , xj = »i xi , xj = »— xi , xj , (A.72)

i

but »— = »i , and so

i

(»i ’ »j ) xi , xj = 0. (A.73)

iv) An operator A is said to be diagonalizable if we can ¬nd a basis for V

that consists of eigenvectors of A. In this basis, A is represented by the

matrix A = diag (»1 , »2 , . . . , »n ), where the »i are the eigenvalues.

Not all linear operators can be diagonalized. The key element determining

the diagonalizability of a matrix is the minimal polynomial equation obeyed

by the matrix representing the operator. As mentioned in the previous sec-

tion, the possible eigenvalues an n — n matrix A are given by the roots of

the characteristic equation

0 = det (A ’ »I) = (’1)n »n ’ tr (A)»n’1 + · · · + (’1)n det (A) .

This is because a non-trivial solution to the equation

Ax = »x (A.74)

requires the matrix A’»I to have a non-trivial nullspace, and so det (A ’ »I)

must vanish. Now Cayley™s Theorem, which we proved in the previous sec-

tion, asserts that every matrix obeys its own characteristic equation:

An ’ tr (A)An’1 + · · · + (’1)n det (A)I = 0.

The matrix A may, however, satisfy an equation of lower degree.

Example: The characteristic equation of the matrix

»1 0

A= (A.75)

0 »1

is (» ’ »1 )2 . Cayley therefore asserts that (A ’ »1 I)2 = 0. This is clearly

true, but A also satis¬es the equation of ¬rst degree (A ’ »1 I) = 0.

Worked Exercise: Suppose that A is hermitian with respect to a positive def-

inite inner product , . Show that the minimal equation has no repeated

roots.

A.6. DIAGONALIZATION AND CANONICAL FORMS 299

Solution: Suppose A has minimal equation (A ’ »I)2 Q = 0 where Q is a

polynomial in A. Then, for all vectors x we have

0 = Qx, (A ’ »I)2 Qx = (A ’ »I)Qx, (A ’ »I)Qx . (A.76)

Now the vanishing of the rightmost expression shows that 0 = (A ’ »I)Qx