Agoston M.K. Computer Graphics and Geometric Modelling: Mathematics

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and

The existence part of the theorem is proved. To prove uniqueness, assume that there

is another vector u¢ in V with a(v) = u¢• v. Then (u - u¢)•v = 0 for all v in V. In par-

ticular, letting v = u - u¢, we get that

which implies that u = u¢ and we are done.

Next, assume that V is a vector space and T : V Æ V is a linear transformation.

Given v Œ V, deﬁne a linear functional T

By Theorem 1.8.2, there is a unique vector v*, so that

Deﬁnition. The map

deﬁned by

is called the adjoint of T.

1.8.3. Lemma. The adjoint map T* satisﬁes

for all v, w Œ V.

Proof. By deﬁnition, T(v)•w = T

(v) = w*•v = T*(w)•v.

1.8.4. Lemma. The adjoint map T* is a linear transformation.

Proof. Using Lemma 1.8.3 and the linearity of the dot product, we have that

uvwuvw

uv uw

uvuw

uv w

∑+

()

∑+

()

∑+

()

∑

=∑

()

+∑

()

=∑

()

()()

Ta b T a b

aT bT

**.

TTvwv w

()

∑=∑

()

T* *vv

()

T*:VVÆ

wvw

()

=∑*.

wwv

()

∑ .

uu uu-¢

()

∑-¢

()

= 0,

uv u x u uu u∑=∑ +

()

=∑=

()

cc ca

aa a avxu u u

()

ccc

1.8 Principal Axes Theorems 39

Since this holds for all vectors u, we must have T*(av + bw) = aT*(v) + bT*(w), which

proves the lemma.

Deﬁnition. A linear transformation T:V Æ V is called self-adjoint if T = T*.

It follows from Lemma 1.8.3 that if a linear transformation T is self-adjoint

then

One can prove the converse, namely

1.8.5. Lemma. If a linear transformation T:V Æ V satisﬁes

for all v, w Œ V, then T is self-adjoint.

Proof. The lemma follows from the fact that for all vectors v we have

1.8.6. Theorem. Let V be a real vector space and T: V Æ V a linear transformation.

If M is the matrix of T with respect to an orthonormal basis, then the matrix of the

adjoint T* of T with respect to that same basis is M

Proof. Let u

, u

,..., u

be an orthonormal basis. By deﬁnition of the matrix for a

linear transformation and properties of orthonormal bases, the ijth entries of the

matrices for T and T* are T(u

)·u

and T*(u

)·u

, respectively. But

1.8.7. Corollary. The matrix for a self-adjoint linear transformation on a real vector

space with respect to an orthonormal basis is symmetric. Conversely, if the matrix for

a linear transformation over a real vector space with respect to an orthonormal basis

is symmetric, then the linear transformation is self-adjoint.

Proof. This is an easy consequence of Theorem 1.8.6.

Self-adjoint transformations are sometimes called symmetric transformations

because of Corollary 1.8.7. The complex analogs of Theorem 1.8.6 and Corollary 1.8.7

simply replace the transpose with the complex conjugate transpose and the self-

adjoint transformations in this case are sometimes called Hermitian.

We now return to the problem of when a linear transformation can be diagonal-

ized. We shall deal with real and complex vector spaces separately. The reason is that

eigenvalues are roots of the characteristic polynomial of a transformation. Although

polynomials always factor completely into linear factors over the complex numbers,

TTT

iji j ji

uuu u uu

()

∑=∑

()

∑**.

vw vwv w∑

()

∑=∑

()

TT T*.

TTvwv w

()

∑=∑

()

TTvwv w

()

∑=∑

()

40 1 Linear Algebra Topics

this is not always the case over the reals. A polynomial may have no roots at all over

the reals.

1.8.8. Lemma. Every eigenvalue of a self-adjoint linear transformation T on a

complex vector space V with inner product • is real.

Proof. Let l be an eigenvalue for T and u a nonzero eigenvector for l. Then

Since u •u π 0, l= , that is, l is real.

1.8.9. Lemma. Let T be a self-adjoint linear transformation over a real n-

dimensional vector space V, n ≥ 1, with inner product •. Then

(1) The characteristic polynomial of T is a product of linear factors.

(2) Eigenvectors corresponding to distinct eigenvalues are orthogonal.

Proof. By passing to the matrix A for T, (1) follows immediately from Lemma 1.8.8

because we can think of A as deﬁning a complex transformation on C

and every poly-

nomial of degree n factors into linear factors over the complex numbers. To prove (2),

assume that T(u) =lu and T(v) =mv for lπm. Then

Since lπm, it follows that u •v = 0, and we are done.

1.8.10. Theorem. (The Real Principal Axes Theorem) Let T be a self-adjoint trans-

formation on an n-dimensional real vector space V, n ≥ 1. Then V admits an ortho-

normal basis u

, u

,..., u

consisting of eigenvectors of T, that is,

for some real numbers l

Proof. The proof is by induction on n. The theorem is clearly true for n = 1. Assume,

therefore, inductively that it has been proved for dimension n - 1, n > 1. There are

basically two steps involved in the rest of the proof.

First, we need to know that the transformation actually has at least one real eigen-

value l. This was proved by Lemma 1.8.9(1). Let v be a nonzero eigenvector for l and

let u

= v/|v|.

The second step, in order to use the inductive hypothesis, is to show that the

orthogonal complement W

of W = <v> = <u

> is an invariant subspace of T. This

follows from the fact that if w Œ W

, then

vw w vwvw∑

()

=∑

()

∑= ∑=TvT T*,l 0

iii

()

ll mmuv uv u v u v u v uv∑

()

=∑=

()

∑=∑

()

=∑ = ∑

()

TT .

ll lluu uu u u u u u u u u uu∑

()

=∑=

()

∑=∑

()

=∑

()

=∑ = ∑

()

TTT*.

1.8 Principal Axes Theorems 41

so that T(w) Œ W

. Clearly, S = T|W

is a self-adjoint transformation on the (n - 1)-

dimensional vector space W

. The inductive hypothesis applied to S means that there

is an orthonormal basis u

, u

,..., u

for W

which are eigenvectors for S (and hence

for T). The vectors u

are obviously what we wanted, proving the theorem.

In the case of Theorem 1.8.10, the name “Principal Axes Theorem” comes from

its role in ﬁnding the principal axes of ellipses. The matrix form of Theorem 1.8.10 is

1.8.11. Theorem. If A is a real symmetric n ¥ n matrix, then there exists an orthog-

onal matrix P so that D = P

-1

AP is a diagonal matrix. In particular, every real sym-

metric matrix is similar to a diagonal one.

Proof. Simply let the columns of P be the vectors that form an orthonormal basis

of eigenvectors.

Note that Theorem 1.8.11 only gives sufﬁcient conditions for a matrix to be similar

to a diagonal one. Nonsymmetric matrices can also be similar to a diagonal one. For

necessary and sufﬁcient conditions for a matrix to be diagonalizable see Theorem

C.4.10.

In Theorem 1.8.11, the number s of positive diagonal entries of D is uniquely deter-

mined by A. We may assume that the diagonal of D has the s positive entries ﬁrst, fol-

lowed by r - s negative entries, followed by n - r zeros, where r is the rank of A.

1.8.12. Example. Let

We want to ﬁnd an orthogonal matrix P so that P

-1

AP is a diagonal matrix.

Solution. Consider A to be the matrix of a linear transformation T on R

. Now, the

roots of the characteristic polynomial

are 1 and 3, which are the eigenvalues of T. To ﬁnd the corresponding eigenvectors,

we must solve

and

This leads to two pairs of equations

-+ =

xy I A

()

xy I A

()

det t I A t t

43-

()

=-+

A =

42 1 Linear Algebra Topics

and

In other words, the vectors v

= (1,1) and v

= (1,-1) are eigenvectors corresponding

to eigenvalues 1 and 3, respectively. Let u

= (1/÷2,1/÷2) and u

= (1/÷2,-1/÷2). If P is

the matrix with columns u

, then

1.8.13. Example. Let

We want to ﬁnd an orthogonal matrix P so that P

-1

AP is a diagonal matrix.

Solution. The roots of the characteristic polynomial

are 1 and 4. To ﬁnd the eigenvectors corresponding to the eigenvalue 1 we need to

solve the equations

The solution set X has the form

Applying the Gram-Schmidt algorithm to the basis (-1,1,0), (-1,0,1) produces the

orthonormal basis u

= (-1/÷2,1/÷2,0) and u

= (-1/÷6,-1/÷6,2/÷6) for X. Next, to ﬁnd

the eigenvector for the eigenvalue 4, we need to solve

xyz

xy z

--=

-+ - =

-- + =.

()

{}

()

{}

y zyz yz y z yz,, , ,, ,, , .RR110 101

-- - =

xyz

det dettI A

21 1

121

11 2

14-

()

---

()

A =

211

121

112

PAP=

and P .

1.8 Principal Axes Theorems 43

The solutions to these equations have the form x(1,1,1). Let u

= (1/÷3,1/÷3,1/÷3).

Finally, if P is the matrix whose columns are the u

, then

Deﬁnition. A linear transformation T is said to be normal if it commutes with its

adjoint, that is, TT* = T*T.

1.8.14. Theorem. (The Complex Principal Axes Theorem) Let T be a normal trans-

formation on an n-dimensional complex vector space V, n ≥ 1. Then V admits an

orthonormal basis u

, u

,..., u

consisting of eigenvectors of T, that is,

for some complex numbers l

Proof. See [Lips68].

The matrix form of Theorem 1.8.14 is

1.8.15. Theorem. If A is a normal matrix, then there exists an unitary matrix P so

that D = P

-1

AP is a diagonal matrix.

Proof. See [Lips68].

1.9 Bilinear and Quadratic Maps

This section describes some maps that appear quite often in mathematics. However,

we are not interested in just the general theory. Quadratic maps and quadratic forms,

in particular, have important applications in a number of areas of geometry and topol-

ogy. For example, the conics, which are an important class of spaces in geometry, are

intimately connected with quadratic forms. Other applications are found in Chapters

8 and 9.

Deﬁnition. A bilinear map on a vector space V over a ﬁeld k is a function

f:V ¥ V Æ k satisfying

(1) f(av + bv¢,w) = af(v,w) + bf(v¢,w), and

(2) f(v,aw + bw¢) = af(v,w) + bf(v,w¢),

for all v, v¢, w, w¢ŒV and a, b Œ k.

iii

()

P and P AP=

100

010

004

44 1 Linear Algebra Topics

1.9.1. Example. The dot product on R

is a bilinear map. More generally, an inner

product is a bilinear map.

1.9.2. Example. The determinant function on R

(where we think of either the rows

or columns of the matrix as vectors in R

) is a bilinear map.

1.9.3. Example. Let A be an n ¥ n matrix. The map f:R

¥ R

Æ R deﬁned by

is a bilinear map.

1.9.4. Example. Let T be a linear transformation on R

. The map f:R

¥ R

Æ R

deﬁned by

is a bilinear map.

Deﬁnition. Let f be a bilinear map on a vector space V. Let B = (v

,...,v

) be an

ordered basis for V and let a

= f(v

). The matrix A = (a

) is called the matrix for f

with respect to the basis B. The determinant of A is called the discriminant of f with

respect to the basis B.

The matrix for a bilinear map clearly depends on the chosen basis. However, the

following is true:

1.9.5. Proposition. If B¢=(v

¢,v

¢,...,v

¢) is another ordered basis for V and if A¢

is the matrix of the bilinear map f with respect to B¢, then

where C = (c

) is the matrix relating the basis B to the basis B¢, that is,

Proof. This can be checked by a straightforward computation.

Deﬁnition. A real n ¥ n matrix A is said to be congruent to a real n ¥ n matrix B if

there exists a nonsingular matrix C such that A = CBC

It is easy to show that the congruence relation is an equivalence relation on the

set of all n ¥ n real matrices. We can rephrase Proposition 1.9.5.

1.9.6. Corollary. The matrix of a bilinear map is unique up to congruence, so that

the study of bilinear maps is equivalent to the study of congruence classes of matrices.

iijj

A CAC

¢= ,

fTv,w v w

()

=∑

()

v,w v w

()

1.9 Bilinear and Quadratic Maps 45

Deﬁnition. The rank of a bilinear map f is the rank of any matrix for f. A bilinear

map on an n-dimensional vector space is said to be degenerate or nondegenerate if its

rank is less than n or equal to n, respectively.

It follows from Proposition 1.9.5 that the rank of a bilinear map is well

deﬁned.

1.9.7. Proposition. The matrix associated to a symmetric bilinear map with respect

to any basis is a symmetric matrix.

Proof. Exercise.

Deﬁnition. A quadratic map on a vector space V over a ﬁeld k is any map q : V Æ k

that can be deﬁned in the form q(v) = f(v,v), where f is some bilinear map on V.

In that case, q is also called the quadratic map associated to f. The quadratic map

q is said to be degenerate or nondegenerate if f is. A discriminant of q is deﬁned to

be discriminant of f with respect to some basis for V. The quadratic map q and the

bilinear map f are said to be positive deﬁnite if q(v) = f(v,v) > 0 for all v π 0.

1.9.8. Example. Let f be a bilinear map on R

and assume that

The quadratic map q associated to f is then given by

We see that q is just a homogeneous polynomial of degree 2 in v

and v

If the ﬁeld k does not have characteristic 2 and if f is a symmetric bilinear map

with associated quadratic map q, then

In other words, knowledge of q alone allows one to reconstruct f, so that the concepts

“symmetric bilinear map” and “quadratic map” are really just two ways of looking at

the same thing.

The form of the quadratic map in Example 1.9.8 and others like it motivates

what is basically nothing but some alternate terminology for talking about quadratic

maps.

Deﬁnition. A d-ic form over a ﬁeld k is a homogeneous polynomial over k of degree

d in an appropriate number of variables. A linear or quadratic form is a d-ic form

where d is 1 or 2, respectively.

For example, 2x + 3y is a linear form in variables x and y and

xyzxyyz

222

523+-++

fqqqvw v w v w,.

()

[]

qavaavvavv

()

=++

()

12 21 1 2 22

favwavwavwavwvw,.

()

=+++

11 1 1 12 1 2 21 2 1 22 2 2

46 1 Linear Algebra Topics

is a quadratic form in variables x, y, and z. Note that the quadratic form can be rewrit-

ten with symmetric cross-terms as

It follows that one can associate the symmetric matrix

with this form. More generally, if the ﬁeld k does not have characteristic 2, such as,

for example, R or C, we can make the cross-terms symmetric with the trick shown in

the example above. It follows that in this case every quadratic form in n variables is

simply an expression of the type

where A = (a

) is a symmetric matrix. This means that every quadratic form deﬁnes

a unique quadratic map

on a vector space V in the following way: Choose an ordered basis B = (v

,...,v

)

for V. Let v Œ V and suppose that

Then

where x = (x

,...,x

). In this case also, the matrix for the associated bilinear map

is just the matrix A. Of course, for all this to make sense we are treating the x

values rather than variables, but we can see that from a theoretical point of view there

is no difference between the theory of quadratic forms and quadratic maps. This

explains why in the literature the terms “quadratic map” and “quadratic form” are

qA axx

ij i j

vxx

()

ÂÂ

vv=

qk: V Æ

axx

ij i j

ÂÂ

xyz xyyxyzzy

222

+-+ + + +.

1.9 Bilinear and Quadratic Maps 47

often used interchangeably. In particular, one uses the same terms, such as “degen-

erate,” “nondegenerate,” “positive deﬁnite,” or “discriminant” for both. One will some-

times also ﬁnd the term “bilinear form” used instead of “bilinear map.”

Note. In this book we shall often use the more popular term “quadratic form” even

though we may interpret it as a quadratic map because that is typically more con-

venient computationally. Speciﬁcally, when the ﬁeld k does not have characteristic 2,

we shall always feel free to switch between a quadratic form and the appropriate cor-

responding quadratic map with its unique associated symmetric bilinear map whose

symmetric matrix is unique up to congruence.

Now, an arbitrary quadratic form can be quite complicated. Key to understand-

ing them is the fact that one can always choose a basis, so that with respect to this

basis, the form has a nice simple structure.

1.9.10. Theorem. (The Principal Axes Theorem) Given a quadratic form q deﬁned

on R

, there exists an orthonormal basis for R

with respect to which q has the

form

The difference s - t is called the signature of the quadratic form or the associated

symmetric bilinear map.

Proof. This is an immediate consequence of Theorem 1.8.11. The integers s and t,

and hence the signature, are independent of the basis and hence invariants of the

quadratic form.

If we do not insist on an orthonormal basis for the diagonalization of a quadratic

form, then there is a weaker version of Theorem 1.9.10. It is interesting because there

is a simpler algorithm for ﬁnding a diagonalizing basis for a quadratic form. Here is

its matrix form.

1.9.11. Theorem. If A is a real symmetric n ¥ n matrix of rank r, then A is con-

gruent to a unique diagonal matrix whose ﬁrst s diagonal entries are +1, the next

r - s entries are -1, and the remaining entries are zeros.

Proof. We sketch a proof. For more details, see [Fink72]. Assume that A is not the

zero matrix; otherwise, there is nothing to prove.

Step 1. To make A congruent to a matrix A

that has a nonzero diagonal element.

If A has a nonzero diagonal element, then let A

= A. If all diagonal elements of

A are zero, let a

be any nonzero entry of A. Let E be the elementary matrix E

(1),

which has 1s on the diagonal, a 1 in the jith place, and zeros everywhere else. Let

= EAE

. The matrix A

is obtained from A by adding the jth row of A to the ith

row followed by adding the jth column of the result to the ith column. It is easy to

see that the ith diagonal element of A

is 2a

and hence nonzero.

q x x x x x x x where

nsss

st st i12 1

0, ,..., ... ... , .

()

=++- -- >

lll l l

48 1 Linear Algebra Topics