If there are no x2 , y 2, or z 2 terms to get us started, then we can proceed by

using (x + y)2 and (x ’ y)2 . For example, consider

Q = 2xy + 2yz + 2zy,

1 1

(x + y)2 ’ (x ’ y)2 + 2xz + 2yz

=

2 2

1 1

(x + y)2 + 2(x + y)z ’ (x ’ y)2

=

2 2

1 1

(x + y + 2z)2 ’ (x ’ y)2 ’ 4z 2

=

2 2

2 2 2

= ξ ’· ’ζ ,

where

1

ξ = √ (x + y + 2z),

2

1

· = √ (x ’ y),

2

√

ζ= 2z.

A judicious combination of these two tactics will reduce the matrix represent-

ing any real quadratic form to a matrix with ±1™s and 0™s on the diagonal,

and zeros elsewhere. As the egregiously asymmetric treatment of x, y, z

in the last example indicates, this can be done in many ways, but Cayley™s

Law of Inertia asserts that the number of +1™s, ’1™s and 0™s will always be

the same. Naturally, if we allow complex numbers in the rede¬nitions of the

variables, we can always reduce the form to one with only +1™s and 0™s.

A.6. DIAGONALIZATION AND CANONICAL FORMS 303

The essential di¬erence between diagonalizing linear maps and diagonal-

izing quadratic forms is that in the former case we seek matrices A such that

A’1 MA is diagonal, whereas in the latter case we seek matrices A such that

AT MA is diagonal. Here, the superscript T denotes transposition.

Exercise: Show that the matrix representing the quadratic form

Q = ax2 + 2bxy + cy 2

may be reduced to

10 1 0 10

, , or ,

0 ’1

01 00

depending on whether the discriminant, ac ’ b2 , is respectively greater than

zero, less than zero, or equal to zero.

Warning: There is no such thing as the determinant of a quadratic form. Of

course you can always compute the determinant of the matrix representing

the quadratic form in some basis, but if you change basis and repeat the

calculation you will get a di¬erent answer.

A.6.3 Symplectic Forms

A skew-symmetric bilinear form ω : V — V ’ R is often called a symplectic

form. Such forms play an important role in Hamiltonian dynamics and in

optics. Let

ω(ei , ej ) = ωij , (A.93)

where ωij compose a real skew symmetric matrix. We will write

1

ω = ωij e—i §, e—j (A.94)

2

2

where the wedge (or exterior ) product, e—j § e—j ∈ (V — ), of a pair of basis

vectors in V — is de¬ned by

ij

e—i § e—j (e± , eβ ) = δ± δβ ’ δβ δ± .

ij

(A.95)

Thus, if x = xi ei and y = y i ei , we have

ω(x, y) = ωij xi y j . (A.96)

304 APPENDIX A. ELEMENTARY LINEAR ALGEBRA

We then extend the de¬nition of the wedge product to other elements of V —

by requiring “§” to be associative and be distributive.

Theorem: For any ω ∈ 2 (V — ) there exists a basis {f —i } of V — such that

ω = f —1 § f —2 + f —3 § f —4 + · · · + f —(p’1) § f —p . (A.97)

Here, the integer p ¤ n is the rank of ω. It is necessarily an even number.

Proof: The proof is a skew-analogue of Lagrange™s method of completing the

square. If

1

ω = ωij e—i § e—j (A.98)

2

is not identically zero, we can, after re-ordering the basis if neceessary, assume

that ω12 = 0. Then

1

ω = e—1 ’ (ω23 e—3 + · · · + ω2n e—n ) §(ω12 e—2 +ω13 e—3 +· · · ω1n e—n )+ω {3}

ω12

(A.99)

where ω {3} ∈ 2 (V — ) does not contain e—1 or e—2 . We set

1

f —1 = e—1 ’ (ω23 e—3 + · · · + ω2n e—n ) (A.100)

ω12

and

f —2 = ω12 e—2 + ω13 e—3 + · · · ω1n e—n . (A.101)

Thus,

ω = f —1 § f —2 + ω {3} . (A.102)

If the remainder ω {3} is identically zero, we are done. Otherwise, we apply

the same same process to ω{3} so as to construct f —3 , f —4 and ω {5} ; we continue

in this manner until we ¬nd a remainder, ω{p+1} , that vanishes.

If {fi } is the basis for V dual to the basis {f —i } then ω(f1 , f2 ) = ’ω(f2 , f1 ) =

ω(f3 , f4 ) = ’ω(f4 , f3 ) = 1, and so on, all other values being zero. Suppose that

we de¬ne the coe¬cients ai j by expressing f —i = ai j e—j , and hence ei = fj aj i .

Then the matrix „¦, with entries ωij , that represents the skew bilinear form

has been expressed as

˜

„¦ = AT „¦A, (A.103)

˜

where A is the matrix with entries ai j , and „¦ is the matrix

«

01

¬ ’1 0

¬ ·

·

˜¬ ·

01

„¦=¬ ·, (A.104)

¬ ·

’1 0

¬ ·

..

.

A.6. DIAGONALIZATION AND CANONICAL FORMS 305

which contains p/2 diagonal blocks of

01

, (A.105)

’1 0

and all other entries are zero.

306 APPENDIX A. ELEMENTARY LINEAR ALGEBRA