(A ’ »I)Q = 0. (A.77)

The equation with the repeated factor was not minimal therefore.

If the equation of lowest degree satis¬ed by the matrix has no repeated

roots, the matrix is diagonalizable; if there are repeated roots, it is not. The

last statement should be obvious, because a diagonalized matrix satis¬es an

equation with no repeated roots, and this equation will hold in all bases,

including the original one. The ¬rst statement, in combination with with

the observation that the minimal equation for a hermitian matrix has no

repeated roots, shows that any hermitian matrix can be diagonalized.

To establish the ¬rst statement, suppose that A obeys the equation

0 = P (A) ≡ (A ’ »1 I)(A ’ »2 I) · · · (A ’ »n I), (A.78)

where the »i are all distinct. Then, setting x ’ A in the identity7

(x ’ »2 )(x ’ »3 ) · · · (x ’ »n ) (x ’ »1 )(x ’ »3 ) · · · (x ’ »n )

+···

1= +

(»1 ’ »2 )(»1 ’ »3 ) · · · (»1 ’ »n ) (»2 ’ »1 )(»2 ’ »3 ) · · · (»2 ’ »n )

(x ’ »1 )(x ’ »2 ) · · · (x ’ »n’1 )

+ , (A.79)

(»n ’ »1 )(»n ’ »2 ) · · · (»n ’ »n’1 )

where in each term one of the factors of the polynomial is omitted in both

numerator an denominator, we may write

I = P 1 + P2 + · · · + P n , (—) (A.80)

where

(A ’ »2 I)(A ’ »3 I) · · · (A ’ »n I)

P1 = , (A.81)

(»1 ’ »2 )(»1 ’ »3 ) · · · (»1 ’ »n )

7

The identity is true because the di¬erence of the left and right hand sides is a poly-

nomial of degree n ’ 1, which, by inspection, vanishes at the n points x = » i . But a

polynomial which has more zeros than its degree, must be identically zero.

300 APPENDIX A. ELEMENTARY LINEAR ALGEBRA

etc. Clearly Pi Pj = 0 if i = j, because the product contains the minimal

equation as a factor. Multiplying (—) by Pi therefore gives P2 = Pi , showing

i

that the Pi are projection operators. Further (A ’ »i I)(Pi ) = 0, so

(A ’ »i I)(Pi x) = 0 (A.82)

for any vector x, and we see that Pi x is an eigenvector with eigenvalue

»i . Thus Pi projects onto the i-th eigenspace. Any vector can therefore be

decomposed

x = P1 x + P2 x + · · · + P n x

= x1 + x2 + · · · + x n , (A.83)

where xi is an eigenvector with eigenvalue »i . Since any x can be written as

a sum of eigenvectors, the eigenvectors span the space.

Jordan Decomposition

If the minimal polynomial has repeated roots, the matrix can still be re-

duced to the Jordan canonical form, which is diagonal except for some 1™s

immediately above the diagonal.

For example, suppose the characteristic equation for a 6 — 6 matrix A is

0 = det (A ’ »I) = (»1 ’ »)3 (»2 ’ »)2 (»3 ’ »), (A.84)

and that this equation is also the minimal polynomial equation. Then the

Jordan form is

«

»1 1 0 0 0 0

¬0 »1 1 0 0 0·

¬

·

¬ ·

¬0 0 »1 0 0 0·

’1

T AT = ¬ ·. (A.85)

¬0 0 0 »2 1 0·

¬ ·

¬ ·

0 0 0 0 »2 0

0 0 0 0 0 »3

One may easily see that the equation above is the minimal equation.

It is rather tedious, but quite straightforward, to show that any linear

map can be reduced to Jordan form. The proof is along the lines of the

example in homework set 0.

A.6. DIAGONALIZATION AND CANONICAL FORMS 301

A.6.2 Quadratic Forms

Do not confuse the notion of diagonalizing the matrix representing a lin-

ear map A : V ’ V with that of diagonalizing the matrix representing a

quadratic form. A (real) quadratic form is a map Q : V ’ R, which is

obtained from a symmetric bilinear form B : V — V ’ R by setting the two

arguments, x and y, in B(x, y) equal:

Q(x) = B(x, x). (A.86)

No information is lost by this specialization. We can recover the non-diagonal

(x = y) values of B from the diagonal values, Q(x), by using the polarization

trick

1

B(x, y) = [Q(x + y) ’ Q(x) ’ Q(y)]. (A.87)

2

An example of a real quadratic form is the kinetic energy term

1 1

T (x) = mij xi xj = x · Mx

™ ™

™ ™™ (A.88)

2 2

in a “small vibrations” Lagrangian. Here, M, with entries mij , is the mass

matrix.

Whilst one can diagonalize such forms by the tedious procedure of ¬nding

the eigenvalues and eigenvectors of the associated matrix, it is simpler to use

Lagrange™s method, which is based on repeatedly completing squares.

Consider, for example, the quadratic form

« «

1 ’2

1 x

2 2 2 ¬ ·¬ ·

Q = x ’ y ’ z + 2xy ’ 4xz + 6yz = ( x, y, z ) 1 ’1 3y .

’2 3 ’1 z

(A.89)

We complete the square involving x:

Q = (x + y ’ 2z)2 ’ 2y 2 + 10yz ’ 5z 2 , (A.90)

where the terms outside the squared group no longer involve x. We now

complete the square in y:

√ 5 15

Q = (x + y ’ 2z) ’ ( 2y ’ √ z)2 + z 2 ,

2

(A.91)

2

2

302 APPENDIX A. ELEMENTARY LINEAR ALGEBRA

so that the remaining term no longer contains y. Thus, on setting

ξ = x + y ’ 2z,

√ 5

2y ’ √ z,

·=

2

15

ζ= z,

2

we have « «

1 00 ξ

2 2 2 ¬ ·¬ ·

Q = ξ ’ · + ζ = ( ξ, ·, ζ ) 0 ’1 0 · . (A.92)