lesson: bilinear, quadratic and hermitian forms lesson ... · 6. quadratic forms on ℝ 7....

Bilinear, Quadratic and Hermitian Forms

Institute of Lifelong Learning, University of Delhi pg. 1

Lesson: Bilinear, Quadratic and Hermitian Forms Lesson Developer: Vivek N Sharma

College / Department: Department of Mathematics, S.G.T.B. Khalsa College, University of Delhi



Table of Contents

1. Learning Outcomes

2. Introduction

3. Definition of a Bilinear Form on a Vector Space

4. Various Types of Bilinear Forms

5. Matrix Representation of a Bilinear Form on a Vector Space

6. Quadratic Forms on ℝ𝑛

7. Hermitian Forms on a Vector Space

8. Summary

9. Exercises

10. Glossary and Further Reading

11. Solutions/Hints for Exercises

1. Learning Outcomes:

After studying this unit, you will be able to

define the concept of a bilinear form on a vector space.

explain the equivalence of bilinear forms with matrices.

represent a bilinear form on a vector space as a square matrix.

define the concept of a quadratic form on ℝ𝑛.

explain the notion of a Hermitian form on a vector space.



2. Introduction:

Bilinear forms occupy a unique place in all of mathematics. The study of linear

transformations alone is incapable of handling the notions of orthogonality in geometry,

optimization in many variables, Fourier series and so on and so forth. In optimization

theory, the relevance of quadratic forms is all the more. The concept of dot product is a

particular instance of a bilinear form. Quadratic forms, in particular, play an all important

role in deciding the maxima-minima of functions of several variables. Hermitian forms

appear naturally in harmonic analysis, communication systems and representation theory.

The theory of quadratic forms derives much motivation from number theory. In short, there

are enough reasons to undertake a basic study of bilinear and quadratic forms.

We first remark that in this lesson, we shall deal with the fields ℱ = ℚ,ℝ,ℂ only. That is, for

our purposes, 𝑐𝑕𝑎𝑟(ℱ) ≠ 2. Now, let us begin with definitions and examples.

3. Definition of a Bilinear Form on a Vector Space:

We know that a linear functional is a scalar-valued linear transformation on a vector space.

In a similar spirit, a bilinear form on a vector space is also a scalar-valued mapping of the

vector space. The difference lies in the fact that while a linear functional is a function of a

single vector variable, a bilinear form is a function of two vector variables. In other words,

while a linear functional on a vector space 𝑉 has the domain set 𝑉, a bilinear form on 𝑉 has

the domain set the Cartesian product 𝑉 × 𝑉. A bilinear form is linear in both the variables.

Hence, the name bears the adjective ‘bilinear’.

3.1. Definition: Let 𝑉 be a vector space over a field ℱ. A bilinear form on a vector space 𝑉

over a field ℱ is a mapping 𝑓:𝑉 × 𝑉 → ℱ such that for all 𝑣1 , 𝑣2, 𝑣,𝑤 𝜀 𝑉 and 𝛼 𝜀 ℱ, we have,

1. 𝑓 𝑣1 + 𝑣2,𝑤 = 𝑓 𝑣1 ,𝑤 + 𝑓 𝑣2 ,𝑤 ;

2. 𝑓 𝑣,𝑤1 + 𝑤2 = 𝑓 𝑣,𝑤1 + 𝑓 𝑣,𝑤2 ;

3. 𝑓 𝛼𝑣,𝑤 = 𝛼𝑓 𝑣,𝑤 ;

4. 𝑓 𝑣,𝛼𝑤 = 𝛼𝑓 𝑣,𝑤 .

Thus, a bilinear form on a vector space 𝑉 is a function on 𝑉 × 𝑉 such that it is linear in both

coordinates.

I.Q.1



3.2. An Important Example of a Bilinear Form:

Every square matrix, having entries from a field ℱ (=ℝ or ℂ), gives rise to a bilinear form. Let

𝐴 be an 𝑛 × 𝑛 matrix over a field ℱ. Then, the function

𝑓:ℱ𝑛 × ℱ𝑛 → ℱ

defined by

𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴𝑦

is a bilinear form, on the vector space 𝑉 = ℱ𝑛 , for every pair of vectors 𝑥, 𝑦 𝜀 ℱ𝑛 . One can

easily verify the bilinearity of the mapping 𝑓 using simple properties of matrix addition,

matrix multiplication and matrix transpose. This example demonstrates that every square

matrix over a field produces a bilinear form. Mere demonstration is not enough. In the

section 5, we shall prove that every square matrix over a field determines a bilinear form.

Let us now look at a specific instance of the mapping 𝑓 just defined.

Example 1: Let ℱ = ℚ and let 𝐴 𝜀 𝐺𝑙3(ℚ) be the matrix 𝐴 =

1 0 2

0 0 3

1 0 0

. Construct the

corresponding ℚ-bilinear form on ℚ3.

Solution: The desired bilinear form 𝑓:ℚ3 × ℚ3 → ℚ is

𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴𝑦 = u v w

1 0 2

0 0 3

1 0 0

r

s

t

= 𝑢𝑟 + 3𝑣𝑡 − 𝑤𝑟 + 2𝑢𝑡;

for all 𝑥 =

u

v

w

𝜀 ℚ3 and 𝑦 =

r

s

t

𝜀 ℚ3.

I.Q.2

Just as there are various types of matrices (like symmetric, diagonal, upper-triangular,

skew-symmetric), there are various kinds of bilinear forms. We shall now study them.



4. Special Types of Bilinear Forms:

Bilinear forms of significant importance include: symmetric, skew-symmetric, and

alternating bilinear forms. The forms are conceptually inter-linked. We begin with their

definitions.

Definitions of Symmetric, Skew-Symmetric and Alternating Bilinear Forms: A

bilinear form 𝑓:𝑉 × 𝑉 → ℱ is symmetric if

𝑓 𝑢, 𝑣 = 𝑓 𝑣,𝑢 ∀ 𝑢, 𝑣 𝜀 𝑉;

skew-symmetric if

𝑓 𝑢, 𝑣 = −𝑓 𝑣,𝑢 ∀ 𝑢, 𝑣 𝜀 𝑉;

and alternating if

𝑓 𝑣, 𝑣 = 0 ∀ 𝑣 𝜀 𝑉.

4.1 Examples and Non-Examples of Symmetric, Skew-Symmetric and

Alternating Bilinear Forms:

(1) The usual dot product of vectors in ℝ𝑛 defines a symmetric bilinear form on ℝ𝑛. The

mapping

𝑓: ℝ𝑛 × ℝ𝑛 ⟶ℝ

defined by

𝑓 𝑥, 𝑦 = 𝑥. 𝑦 = 𝑥1𝑦1 + 𝑥2𝑦2 + ⋯+ 𝑥𝑛𝑦𝑛

is a bilinear form on ℝ𝑛 for every vector 𝑥 = (𝑥𝑖)𝑖=1 1 𝑛 and 𝑦 = (𝑦𝑖)𝑖=1 1 𝑛

in ℝ𝑛 and satisfies the symmetry property

𝑓 𝑥, 𝑦 = 𝑓(𝑦, 𝑥),

since the dot product is commutative.

(2) Let 𝑉 = ℝ2 and its elements be viewed as column vectors. Then, the determinant map

𝑑𝑒𝑡: ℝ2 × ℝ2 ⟶ ℝ

given by

𝑑𝑒𝑡 a

b

,c

d

= 𝑑𝑒𝑡a c

b d

= 𝑎𝑑 − 𝑏𝑐

is a skew-symmetric and alternating bilinear form on ℝ2. Skew-symmetry follows

because interchanging the two columns of the matrix changes the sign of the

determinant; and it is alternating because whenever the two columns are identical, the

determinant is zero.



(3) The bilinear form 𝑓:ℂ × ℂ ⟶ ℂ given by

𝑓 𝑧,𝑤 = 𝐼𝑚 𝑧𝑤 ∀ 𝑧,𝑤 𝜀 ℂ

is a skew-symmetric bilinear form on ℂ because, ∀ 𝑧,𝑤 𝜀 ℂ, we have

𝑓 𝑧,𝑤 = 𝐼𝑚 𝑧𝑤 = −𝐼𝑚 𝑤𝑧 = −𝑓(𝑤, 𝑧).

(4) The bilinear form 𝑓:ℚ3 × ℚ3 → ℚ (example 1) given by

𝑓 𝑥, 𝑦 = = 𝑢𝑟 + 3𝑣𝑡 − 𝑤𝑟 + 2𝑢𝑡 for all 𝑥 =

u

v

w

𝜀 ℚ3 and 𝑦 =

r

s

t

𝜀 ℚ3 is neither

symmetric nor alternating on ℚ3.

Value Addition : Remarks

In all characteristics, an alternating bilinear form is skew-symmetric. However, in case

of characteristic not 2, a bilinear form is skew-symmetric if, and only if, it is

alternating.

In characteristic not 2, every bilinear form 𝑓 𝑢, 𝑣 on 𝑉 can be written uniquely as a

sum of a symmetric and an alternating bilinear form, as for every 𝑢, 𝑣 𝜀 𝑉, we have

𝑓 𝑢, 𝑣 = 𝑓 𝑢 ,𝑣 +𝑓(𝑣,𝑤)

2+

𝑓 𝑢 ,𝑣 −𝑓(𝑣,𝑤)

2.

In characteristic not 2, every symmetric bilinear form 𝑓 𝑢, 𝑣 on an 𝑛-dimensional

vector space 𝑉 is completely determined by its values 𝑓 𝑣, 𝑣 on the diagonal, as for

every 𝑢, 𝑣 𝜀 𝑉, we have, using the symmetry of the bilinear form 𝑓,

𝑓 𝑢, 𝑣 = 𝑓 𝑢+𝑣,𝑣+𝑣 −𝑓 𝑣,𝑣 −𝑓(𝑢 ,𝑢)

2=

1

2 𝑓 𝑢, 𝑣 + 𝑓 𝑣,𝑢 = 𝑓(𝑢, 𝑣).

This is the famous polarization identity.



5. Matrix Representation of a Bilinear Form on a Vector Space:

We have seen in section 3 that every square matrix over a field gives rise to a bilinear form.

So, it is natural to ask the converse: does a bilinear form on a vector space also produce a

matrix? In the context of finite dimensional vector spaces, the answer is YES. Every bilinear

form on a finite dimensional vector space can be represented as a matrix. This is what we

shall now learn. Firstly, we define the notion of a matrix of a bilinear form on a finite

dimensional vector space over a field.

Definition of Matrix of a Bilinear Form: Let 𝑉 be an 𝑛-dimensional vector space over a

field ℱ. Let 𝛽 = {𝑒1, 𝑒2,… , 𝑒𝑛} be an ordered basis for 𝑉 over ℱ. Then, the matrix of 𝑓 relative

to the ordered basis 𝛽 is the matrix

𝐴 = [𝑎𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛

defined by

𝑎𝑖𝑗 = 𝑓 𝑒𝑖 , 𝑒𝑗 ∀ 𝑖, 𝑗 = 1 1 𝑛.

This matrix is also known as inner product matrix of 𝑓 relative to the ordered basis 𝛽.

Now, let us look at some explicit examples.

Example 2: Calculate the matrix of the bilinear form 𝑑𝑒𝑡 on ℝ2 relative to the standard

basis for ℝ2.

Solution: The bilinear form 𝑓 = 𝑑𝑒𝑡 on ℝ2 is given by

𝑓 a

b

,c

d

= 𝑑𝑒𝑡 a

b

,c

d

= 𝑑𝑒𝑡a c

b d

= 𝑎𝑑 − 𝑏𝑐;

so that with respect to the standard basis 𝛽 = 𝑒1 =1

0

, 𝑒2 =0

1

, we have,



𝑎11 = 𝑓 𝑒1 , 𝑒1 = 𝑑𝑒𝑡1 1

0 0

= 0; 𝑎12 = 𝑓 𝑒1, 𝑒2 = 𝑑𝑒𝑡1 0

0 1

= 1;

𝑎21 = 𝑓 𝑒2, 𝑒1 = 𝑑𝑒𝑡0 1

1 0

= −1; 𝑎22 = 𝑓 𝑒2, 𝑒2 = 𝑑𝑒𝑡0 0

1 1

= 0;

and therefore, the matrix of this bilinear form 𝑑𝑒𝑡 is 𝐴 =0 1

1 0

.

Example 3: Compute the matrix of the ℚ-bilinear form 𝑓:ℚ3 × ℚ3 → ℚ given by 𝑓 𝑥, 𝑦 =

𝑥1𝑦2 + 𝑥3𝑦2 + 𝑥2𝑦1 , ∀ 𝑥 = 𝑥1, 𝑥2 , 𝑥3 , 𝑦 = 𝑦1 , 𝑦2 , 𝑦3 𝜀 ℚ3 relative to the standard basis for ℚ3.

Solution: The standard basis for ℚ3 is the usual basis

𝛽 = 𝑒1 = 1, 0, 0 , 𝑒2 = 0, 1, 0 , 𝑒3 = (0, 0, 1) ;

and therefore, we have,

𝑎11 = 𝑓 𝑒1 , 𝑒1 = 0; 𝑎12 = 𝑓 𝑒1, 𝑒2 = 1; 𝑎13 = 𝑓 𝑒1, 𝑒3 = 0;

𝑎21 = 𝑓 𝑒2, 𝑒1 = 1; 𝑎22 = 𝑓 𝑒2, 𝑒2 = 0; 𝑎23 = 𝑓 𝑒2, 𝑒3 = 0;

𝑎31 = 𝑓 𝑒3, 𝑒1 = 0; 𝑎32 = 𝑓 𝑒3, 𝑒2 = 1; 𝑎33 = 𝑓 𝑒3, 𝑒3 = 0.

Hence, the matrix of this bilinear form 𝑓 is 𝐴 =

0 1 0

1 0 0

0 1 0

.

I.Q.3

We shall now formally prove that every square matrix is a matrix of a bilinear form; and

that every bilinear form is completely determined by its matrix.

Theorem 1: Let 𝑉 be an 𝑛-dimensional vector space over a field ℱ.

(i) Every 𝑛 × 𝑛 matrix 𝐴 over ℱ is the matrix of some bilinear form 𝑓 on 𝑉 relative to some

ordered basis of 𝑉.

(ii) The matrix of any bilinear form 𝑓 on 𝑉 relative to any ordered basis of 𝑉 completely

determines 𝑓 on 𝑉.

Proof: The proofs for both the parts are separate.

(i) Let 𝐴 = [𝑎𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛 be any 𝑛 × 𝑛 matrix over ℱ. The mapping

𝑓:𝑉 × 𝑉 → ℱ

defined by



𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴𝑦 = 𝑥1 , 𝑥2,… , 𝑥𝑛 .𝐴.

2

1

.

.

.

n

y

y

y

defines a bilinear form on 𝑉. Now, let

𝛽 = 𝑒𝑖 = 0, 0,… , 1𝑖 ,… , 0 𝑖 = 1 1 𝑛}

be the canonical basis for 𝑉 over ℱ. Then, we can see that, ∀ 𝑖, 𝑗 = 1 1 𝑛,

𝑓 𝑒𝑖 , 𝑒𝑗 = 𝑒𝑖𝑡 .𝐴. 𝑒𝑗

= 0, 0,… , 1𝑖 ,… , 0

. . . . .

. . . . .

. . . .

. . . . .

. . . . .

ija

0

.

.

.

1

.

.

.

0

j

= 𝑎𝑖𝑗

and therefore, the matrix 𝐵 of 𝑓 is 𝐴 itself:

𝐵 = [𝑎𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛 = 𝐴.



Thus, the square 𝑛 × 𝑛 matrix 𝐴 is the matrix of the bilinear form 𝑓 relative to the

standard basis of 𝑉 over ℱ. Since our choice of the matrix 𝐴 was arbitrary, the

assertion holds good in the general case. The proof is now complete.

(ii) Let the square 𝑛 × 𝑛 matrix 𝐴 given by

𝐴 = [𝑎𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛

be the matrix of a bilinear form 𝑓:𝑉 × 𝑉 → ℱ relative to a basis

𝛽 = 𝑣1 , 𝑣2,… , 𝑣𝑛

for 𝑉 over ℱ. Thus, ∀ 𝑖, 𝑗 = 1 1 𝑛, we have,

𝑎𝑖𝑗 = 𝑓(𝑣𝑖 ,𝑣𝑗 ). (A)

We want to show that the value of this bilinear form 𝑓(𝑥, 𝑦) at any pair of vectors

𝑥, 𝑦 𝜀 𝑉 can be obtained from the entries of the matrix 𝐴. Since 𝛽 is a basis for 𝑉 over ℱ

and 𝑥, 𝑦 𝜀 𝑉, there exist scalars 𝑥𝑖 , 𝑦𝑖 𝜀 ℱ for 𝑖 = 1 1 𝑛, such that,

𝑥 =1

n

i

i ix v

and 𝑦 =1

n

i

i iy v

(B)

so that, we have,

𝑓 𝑥, 𝑦 = 𝑓(1

n

i

i ix v

, 𝑦) (By (B))

=1

( , )i i

n

i

x f v y

(By linearity in first variable)

=1 1

( , )n n

j j

i j

i ix f v y v

(By (B))

=1 1

( , )n n

ji j

i j

ix y f v v

(By linearity in second variable)



=1 1

n n

j

i j

ix y

𝑎𝑖𝑗 (By (A));

and thus, knowing the matrix of a bilinear form relative to any basis completely determines

the bilinear form. This completes the proof.

The moral of the above theorem is that the nature of the matrix reveals a lot about the

behavior of the corresponding bilinear form.

Now, a natural follow up question would be: what is the impact of basis change on the

matrix representation of a bilinear form? The answer is contained in the following beautiful

theorem.

Theorem 2: Let 𝑉 be an 𝑛-dimensional vector space over a field ℱ. Let 𝐴 be the matrix of a

bilinear form 𝑓:𝑉 × 𝑉 → ℱ relative to some ordered basis 𝛽 for 𝑉 over ℱ. If 𝛽′ is another

ordered basis for 𝑉 over ℱ, then the matrix representation of 𝑓 relative to the basis 𝛽′ is

𝑃𝑡𝐴𝑃 where 𝑃 is the transition matrix describing the basis change 𝛽′ ⟶ 𝛽.

Proof: Let 𝛽 = 𝑣1 , 𝑣2 ,… , 𝑣𝑛 and 𝛽′ = 𝑣1′, 𝑣2′,… , 𝑣𝑛 ′ . Since 𝑃 = [𝑝𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛 is the transition

matrix describing the basis change 𝛽′ ⟶ 𝛽, we have,

𝑣𝑗′ =

1

n

i

ij ip v

∀ 𝑗 = 1 1 𝑛. (A)

Value Addition : Remarks

A bilinear form is symmetric if, and only if, its matrix relative to some basis is

symmetric.

A bilinear form is alternating if, and only if, its matrix relative to some basis is skew-

symmetric.



Hence, the new matrix 𝐴′ = [𝑎𝑖𝑗 ′]𝑖 ,𝑗=1 1 𝑛 representing the bilinear form relative the ordered

basis 𝛽′ is given by

𝑎𝑖𝑗′ = 𝑓(𝑣𝑖

′ , 𝑣𝑗′)

= 𝑓(1

n

k

k

kip v

,1

n

l

l

ljp v

) (By (A))

=1 1

n n

ki

k l

ljp p

𝑓 𝑣𝑘 , 𝑣𝑙 (By the bilinearity of 𝑓)

=1 1

n n

ki

k l

ljp p

𝑎𝑘𝑙 (Because 𝑎𝑘𝑙 = 𝑓 𝑣𝑘 , 𝑣𝑙 ∀ 𝑘, 𝑙 = 1 1 𝑛);

that is,

𝑎𝑖𝑗′ =

1 1

kl l

n n

ki

l

j

k

p pa

so that, we get, ∀ 𝑖, 𝑗 = 1 1 𝑛,

𝑎𝑖𝑗′ =

1 1

kl l

n n

ik

l

j

k

p pa

𝑡

confirming the theorem that

𝐴′ = 𝑃𝑡𝐴𝑃.

The proof is now complete.

Let us see this theorem in actual practice.



Example 4: Consider the ℝ-bilinear form on ℝ3, relative to the standard basis, given by

𝑥, 𝑦 = 2𝑥1𝑦1 − 3𝑥2𝑦2 + 𝑥3𝑦3 ∀ 𝑥 = 𝑥1 , 𝑥2 , 𝑥3 , 𝑦 = 𝑦1 , 𝑦2 , 𝑦3 𝜀 ℝ3. Determine the matrix

representation of the equivalent form relative to the ordered basis 𝛽′ = {𝑣1 = 1, 1, 1 , 𝑣2 =

−2, 1, 1 , 𝑣3 = 2, 1, 0 }.

Solution: We first calculate the matrix 𝐴 representing the given bilinear form relative to the

standard basis for ℝ3. We are given that

𝑓 𝑥, 𝑦 = 2𝑥1𝑦1 − 3𝑥2𝑦2 + 𝑥3𝑦3 ∀ 𝑥 = 𝑥1 , 𝑥2 , 𝑥3 , 𝑦 = 𝑦1 , 𝑦2 , 𝑦3 𝜀 ℝ3,

from which we obtain

𝑎11 = 𝑓 𝑒1 , 𝑒1 = 2; 𝑎12 = 𝑓 𝑒1, 𝑒2 = 0; 𝑎13 = 𝑓 𝑒1, 𝑒3 = 0;

𝑎21 = 𝑓 𝑒2, 𝑒1 = 0; 𝑎22 = 𝑓 𝑒2, 𝑒2 = 0; 𝑎23 = 𝑓 𝑒2, 𝑒3 = −3;

𝑎31 = 𝑓 𝑒3, 𝑒1 = 0; 𝑎32 = 𝑓 𝑒3, 𝑒2 = 1; 𝑎33 = 𝑓 𝑒3, 𝑒3 = 1.

Hence, the matrix of this bilinear form 𝑓 relative to the standard basis for ℝ3 is 𝐴 =

2 0 0

0 0 3

0 0 1

.

Now, the matrix 𝑃 describing the basis change 𝛽′ ⟶ 𝛽 is 𝑃 =

1 2 2

1 1 1

1 1 0

.

Hence, the required matrix 𝐴′ for the new equivalent bilinear form is given by

𝐴′ = 𝑃𝑡𝐴𝑃

=

1 2 2

1 1 1

1 1 0

𝑡

2 0 0

0 0 3

0 0 1

1 2 2

1 1 1

1 1 0

=

1 1 1

2 1 1

2 1 0

2 0 0

0 0 3

0 0 1

1 2 2

1 1 1

1 1 0

=

0 6 4

6 6 8

1 11 8

.



6. Quadratic Forms on ℝ𝒏:

Symmetric matrices and symmetric bilinear forms are fascinating in their own right. Both

are equivalent and arise naturally in various contexts. Symmetric bilinear forms give rise to

what are known as quadratic forms. In optimization theory, machine learning, applied

probability, and above all, in number theory, quadratic forms have deep and serious

applications. There are many open problems regarding quadratic forms.

Let us begin with the definition. In this section, we take ℱ = ℝ.

Definition of a Quadratic Form on a Vector Space: Let 𝑉 be an 𝑛-dimensional vector

space over ℝ. Let 𝑓 be a symmetric bilinear form on 𝑉 with the matrix 𝐴 relative to the

standard basis for 𝑉. The mapping 𝑞:𝑉 → ℝ defined by

𝑞 𝑥 = 𝑓 𝑥, 𝑥 = 𝑥𝑡𝐴𝑥 ∀ 𝑥 𝜀 𝑉

is called a quadratic form on the vector space 𝑉. Equivalently, we say that an expression of

the form

𝑞 𝑥 =1 1

n

ij i j

n

ji

a x x

∀ 𝑥 = (𝑥1 , 𝑥2,… , 𝑥𝑛) 𝜀 𝑉

defines a quadratic form 𝑞:𝑉 → ℝ on the vector space 𝑉.

There is a special way to write the matrix for a quadratic form. We illustrate it through

examples.



Example 5: Determine the matrix of the quadratic form 𝑞:ℝ2 → ℝ given by 𝑞 𝑥, 𝑦 = 𝑥2 − 𝑥𝑦 +

𝑦2.

Solution: We write the quadratic form

𝑞 𝑥, 𝑦 = 𝑥2 − 𝑥𝑦 + 𝑦2

as

𝑞 𝑥, 𝑦 = 𝑥𝑥 −1

2𝑥𝑦 −

1

2𝑦𝑥 + 𝑦𝑦

and hence the matrix of this quadratic form is

𝐴 =

11

2

11

2

.

We may verify that ∀ 𝑢 = 𝑥, 𝑦 𝜀 ℝ2,

𝑢𝑡𝐴𝑢 = x y

11

2

11

2

x

y

= 𝑥2 − 𝑥𝑦 + 𝑦2 = 𝑞(𝑥, 𝑦).

Example 6: Determine the matrix of the quadratic form 𝑞:ℝ2 → ℝ given by 𝑞 𝑥, 𝑦 = 49𝑥2 −

72𝑥𝑦 + 121𝑦2.


𝑞 𝑥, 𝑦 = 49𝑥2 − 72𝑥𝑦 + 121𝑦2

as

𝑞 𝑥, 𝑦 = 49. 𝑥𝑥 − 36. 𝑥𝑦 − 36. 𝑦𝑥 + 121𝑦𝑦

and hence the matrix of this quadratic form is



𝐴 =49 36

36 121

.

We may verify that ∀ 𝑢 = 𝑥, 𝑦 𝜀 ℝ2,

𝑢𝑡𝐴𝑢 = x y49 36

36 121

x

y

= 49𝑥2 − 72𝑥𝑦 + 121𝑦2 = 𝑞(𝑥, 𝑦).

Example 7: Determine the matrix of the quadratic form 𝑞:ℝ3 → ℝ given by 𝑞 𝑥, 𝑦, 𝑧 = 𝑥2 +

2𝑦2 − 5𝑧2 − 𝑥𝑦 + 4𝑦𝑧 − 3𝑧𝑥.


𝑞 𝑥, 𝑦, 𝑧 = 𝑥2 + 2𝑦2 − 5𝑧2 − 𝑥𝑦 + 4𝑦𝑧 − 3𝑧𝑥

as

𝑞 𝑥, 𝑦, 𝑧 = 𝑥𝑥 + 2𝑦𝑦 − 5𝑧𝑧 −1

2𝑥𝑦 −

1

2𝑦𝑥 + 2𝑦𝑧 + 2𝑧𝑦 −

3

2𝑧𝑥 −

3

2𝑥𝑧

and rearranging ‘row-wise’, we get

𝑞 𝑥, 𝑦, 𝑧 = 𝑥𝑥 −1

2𝑥𝑦 −

3

2𝑥𝑧 −

1

2𝑦𝑥 + 2𝑦𝑦 + 2𝑦𝑧 −

3

2𝑧𝑥 + 2𝑧𝑦 − 5𝑧𝑧.

Hence, the matrix of this quadratic form is

𝐴 =

1 31

2 2

12 2

2

32 5

2

.

We may verify that ∀ 𝑢 = 𝑥, 𝑦, 𝑧 𝜀 ℝ3,



𝑢𝑡𝐴𝑢 = x y z

1 31

2 2

12 2

2

32 5

2

x

y

z

= 𝑥2 + 2𝑦2 − 5𝑧2 − 𝑥𝑦 + 4𝑦𝑧 − 3𝑧𝑥

= 𝑞(𝑥, 𝑦, 𝑧).

6.1 Canonical way of writing a quadratic form:

How can we simplify complicated quadratic forms into the simple, manageable ones? This

question was satisfactorily handled by Sylvester and Lagrange in 19th century. Their

contributions are contained in the following theorems.

Theorem 3: Every square matrix of order 𝑛 is congruent to a unique matrix of the form

𝐵 =

r

s

I

I

O

.

Theorem 4 (Sylvester): Let 𝑉 be an 𝑛-dimensional vector space over ℝ endowed with a

quadratic form 𝑞. Then, there exists an ordered basis 𝛽 = 𝑣1 , 𝑣2 ,… , 𝑣𝑛 of 𝑉 such that ∀ 𝑥 𝜀 𝑉,

we have,

𝑥 =1

n

i i

i

v

⟹ 𝑞 𝑥 = (𝑥1)2 + (𝑥2)2 + ⋯+ (𝑥𝑟)2 − (𝑥𝑟+1)2 −⋯− (𝑥𝑠)2.

The integers 𝑟 and 𝑠 do not depend upon the choice of the basis. Their sum 𝑟 + 𝑠 is called

the rank of the quadratic form and the difference 𝑟 − 𝑠 is called the signature of the

quadratic form. Alternatively, the rank of a quadratic form is also defined to the rank of its

matrix representation.



While Sylvester proved the existence of this canonical form, it was Lagrange who gave a

simple algorithm of ‘completing the squares’ to reduce a quadratic form to its canonical

form. We illustrate the process using examples.

Example 8: Determine the rank and signature of the quadratic form 𝑞:ℝ3 → ℝ given by

𝑞 𝑥, 𝑦, 𝑧 = 4𝑥2 + 10𝑦2 + 11𝑧2 − 4𝑥𝑦 − 12𝑦𝑧 + 12𝑧𝑥 by reducing it to its canonical form.

Solution: We see that

𝑞 𝑥, 𝑦, 𝑧 = 4𝑥2 + 10𝑦2 + 11𝑧2 − 4𝑥𝑦 − 12𝑦𝑧 + 12𝑧𝑥

= 4(𝑥 −1

2𝑦 +

3

2𝑧)2 + 9(𝑦 −

1

3𝑧)2 + 𝑧2

so that setting

𝑢 = 2(𝑥 −1

2𝑦 +

3

2𝑧); 𝑣 = 3(𝑦 −

1

3𝑧); and 𝑤 = 𝑧;

we obtain the desired canonical form as

𝑞 𝑢, 𝑣,𝑤 = 𝑢2 + 𝑣2 + 𝑤2;

and hence, signature is 3. Further, the matrix for the canonical form is

𝐴 =

1 0 0

0 1 0

0 0 1

and therefore, rank is 3.

Example 9: Determine the rank and signature of the quadratic form 𝑞:ℝ3 → ℝ given by

𝑞 𝑥, 𝑦, 𝑧 = 2𝑥𝑦 + 2𝑦𝑧 by reducing it to its canonical form.

Solution: We see that

𝑞 𝑥, 𝑦, 𝑧 = 2𝑥𝑦 + 2𝑦𝑧

= 1

2[ 𝑥 + 𝑦 + 𝑧 2 − (𝑥 − 𝑦 + 𝑧)2]



so that the matrix for the canonical form is 𝐴 =

10 0

2

10 0

2

0 0 0

, and hence the

rank is 2. Further, the signature is zero.

7. Hermitian Forms on a Vector Space:

Signals and systems in electrical engineering heavily depend upon Fourier Analysis. And a

Fourier Transform is a typical example of a Hermitian Form on a complex vector space.

Hermitian forms are defined only for complex vector spaces. They differ from the usual

bilinear forms in a subtle way. We begin with the definition.

Definition of Hermitian Form: A Hermitian form on a complex vector space 𝑉 is a map

<, >:𝑉 × 𝑉 ⟶ ℂ such that ∀ 𝑧1 , 𝑧2,𝑤1 ,𝑤2 𝜀 𝑉 and 𝜇 𝜀 ℂ one has

1. < 𝑧1 + 𝑧2 ,𝑤1 >=< 𝑧1 ,𝑤1 > +< 𝑧2,𝑤1 >;

2. < 𝑧1,𝑤1 + 𝑤2 >=< 𝑧1 ,𝑤1 > +< 𝑧1 ,𝑤2 >;

3. < 𝜇𝑧1,𝑤1 >= 𝜇 < 𝑧1,𝑤1 >;

4. < 𝑧1, 𝜇𝑤1 >= 𝜇 < 𝑧1,𝑤1 >.

7.1 Examples of Hermitian Forms:

(1) On 𝑉 = ℂ𝑛 we have the standard Hermitian form given as

< 𝑧1 , 𝑧2,… , 𝑧𝑛 , 𝑤1 ,𝑤2 ,… ,𝑤𝑛 >= 𝑧1 𝑤1 + 𝑧2 𝑤2 + ⋯+ 𝑧𝑛 𝑤𝑛

for every vector 𝑧1, 𝑧2 ,… , 𝑧𝑛 and 𝑤1 ,𝑤2 ,… ,𝑤𝑛 𝜀 ℂ.

(2) On the vector space of all square 𝑛 × 𝑛 matrices over ℂ, the trace map defines a

Hermitian form:

< 𝐴,𝐵 >= 𝑡𝑟 𝐴∗𝐵 .

At last, we have the matrix representation for Hermitian forms.

7.2 Matrix Representation of Hermitian Forms:

Let 𝑉 be an 𝑛-dimensional complex vector space endowed with a Hermitian form < , >. The

matrix representing this form relative to a basis 𝛽 = 𝑣1 , 𝑣2 ,… , 𝑣𝑛 of 𝑉 is

𝐴 = [𝑎𝑖𝑗 ]𝑖 ,𝑗=1 1 𝑛 , where, 𝑎𝑖𝑗 =< 𝑣𝑖 , 𝑣𝑗 > ∀ 𝑖, 𝑗 = 1 1 𝑛.

I.Q.4



8. Summary:

(i) A bilinear form on a vector space 𝑉 is a function 𝑉 × 𝑉 ⟶ ℱ such that it is linear in

both coordinates.

(ii) Let 𝐴 be an 𝑛 × 𝑛 matrix over a field ℱ. Then, the function 𝑓:ℱ𝑛 × ℱ𝑛 → ℱ

defined by 𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴𝑦 is a bilinear form, on the vector space 𝑉 = ℱ𝑛 , for

every pair of vectors 𝑥, 𝑦 𝜀 ℱ𝑛 .

(iii) A bilinear form 𝑓:𝑉 × 𝑉 → ℱ is symmetric if 𝑓 𝑢, 𝑣 = 𝑓 𝑣,𝑢 ∀ 𝑢, 𝑣 𝜀 𝑉;

skew-symmetric if 𝑓 𝑢, 𝑣 = −𝑓 𝑣,𝑢 ∀ 𝑢, 𝑣 𝜀 𝑉; and alternating if 𝑓 𝑣, 𝑣 = 0 ∀ 𝑣 𝜀 𝑉.

(iv) Polarization Identity: In characteristic not 2, every symmetric bilinear form

𝑓 𝑢, 𝑣 on an 𝑛-dimensional vector space 𝑉 is completely determined by its

values 𝑓 𝑣, 𝑣 on the diagonal, as for every 𝑢, 𝑣 𝜀 𝑉, we have, using the symmetry

of the bilinear form 𝑓,

𝑓 𝑢, 𝑣 = 𝑓 𝑢+𝑣,𝑣+𝑣 −𝑓 𝑣,𝑣 −𝑓(𝑢 ,𝑢)

2=

1

2 𝑓 𝑢, 𝑣 + 𝑓 𝑣,𝑢 = 𝑓(𝑢, 𝑣).

(v) Matrix Representation of a Bilinear Form: Let 𝑉 be an 𝑛-dimensional vector

space over a field ℱ. Then,

(a) Every 𝑛 × 𝑛 matrix 𝐴 over ℱ is the matrix of some bilinear form 𝑓 on 𝑉

relative to some ordered basis of 𝑉.

(b) The matrix of any bilinear form 𝑓 on 𝑉 relative to any ordered basis of 𝑉

completely determines 𝑓 on 𝑉.

(vi) A bilinear form is symmetric if, and only if, its matrix relative to some basis is

symmetric.

(vii) Let 𝑉 be an 𝑛-dimensional vector space over a field ℱ. Let 𝐴 be the matrix of a

bilinear form 𝑓:𝑉 × 𝑉 → ℱ relative to some ordered basis 𝛽 for 𝑉 over ℱ. If 𝛽′ is

another ordered basis for 𝑉 over ℱ, then the matrix representation of 𝑓 relative to

the basis 𝛽′ is 𝑃𝑡𝐴𝑃 where 𝑃 is the transition matrix describing the basis change

𝛽′ ⟶ 𝛽.



(viii) To every square 𝑛 × 𝑛 matrix 𝐴 over ℝ, the mapping 𝑞:𝑉 → ℝ given by 𝑞 𝑥 =

𝑓 𝑥, 𝑥 = 𝑥𝑡𝐴𝑥 ∀ 𝑥 𝜀 𝑉 defines a quadratic form on an 𝑛-dimensional real vector

space 𝑉.

(ix) Every square matrix of order 𝑛 is congruent to a unique matrix of the form

𝐵 =

r

s

I

I

O

.

(x) Sylvester’s Theorem: Let 𝑉 be an 𝑛-dimensional vector space over ℝ endowed

with a quadratic form 𝑞. Then, there exists an ordered basis 𝛽 = 𝑣1 , 𝑣2 ,… , 𝑣𝑛 of 𝑉

such that ∀ 𝑥 𝜀 𝑉, we have,

𝑥 =1

n

i i

i

v

⟹ 𝑞 𝑥 = (𝑥1)2 + (𝑥2)2 + ⋯+ (𝑥𝑟)2 − (𝑥𝑟+1)2 −⋯− (𝑥𝑠)2.

(xi) The integers 𝑟 and 𝑠 do not depend upon the choice of the basis. Their sum 𝑟 + 𝑠

is called the rank of the quadratic form and the difference 𝑟 − 𝑠 is called the

signature of the quadratic form. Alternatively, the rank of a quadratic form is

also defined to the rank of its matrix representation.

(xii) Hermitian forms are defined only for complex vector spaces. A Hermitian form on

a complex vector space 𝑉 is a function < , >:𝑉 × 𝑉 ⟶ ℱ such that it is

linear in one coordinate and conjugate-linear in the other coordinate.



Exercises

1. Construct the bilinear form the following matrices:

(a) 𝐴1 =

1 2 5

2 0 1

0 6 6

(b) 𝐴2 =

3 1 3

2 3 0

0 0 0

2. Determine the matrix of the following bilinear forms relative to the standard basis of

appropriate vector space:

(a) 2𝑥1𝑦1 − 3𝑥1𝑦3 + 2𝑥2𝑦2 − 5𝑥2𝑦3 + 4𝑥3𝑦1

(b) 2𝑥1𝑦1 + 𝑥1𝑦2 + 𝑥1𝑦3 + 3𝑥2𝑦1 − 2𝑥2𝑦3 + 𝑥3𝑦2 − 5𝑥3𝑦3

3. In each of the two cases below, a bilinear form is given relative to the standard basis

of appropriate vector space. Determine the matrix representation relative to the new

ordered basis given for each case:

(a) 𝑥1𝑦1 + 𝑥1𝑦2 + 𝑥2𝑦1 + 𝑥2𝑦2; 𝛽′ = {𝑣1 = 1,−1 , 𝑣2 = 1, 1 }

(b) 2𝑥1𝑦1 + 𝑥2𝑦3 − 3𝑥3𝑦2 + 𝑥3𝑦3; 𝛽′ = {𝑣1 = 1,2,3 , 𝑣2 = −1,1 2 , 𝑣3 = (1,2,1) }

4. Construct the quadratic form the following matrices:

(a) 𝐴3 =

0 5 1

5 1 6

1 6 2

(b) 𝐴4 =

a h g

h b f

g f c

5. Reduce each of the following quadratic forms to their respective canonical forms.

Hence, determine their rank and signature.

(a) 2𝑦2 − 𝑧2 + 𝑥𝑦 + 𝑥𝑧

(b) 𝑦𝑧 + 𝑥𝑧 + 𝑥𝑦 + 𝑥𝑡 + 𝑦𝑡 + 𝑧𝑡



Glossary

Bilinear form, symmetric bilinear form, alternating bilinear form, matrix of a bilinear form,

quadratic form, rank, signature, Hermitian form

Further Reading

1. Linear Algebra (2/e) by Kenneth Hoffman and Ray Kunze

Publication Year: 1971

Publisher: Prentice Hall, Inc.

ISBN: 978-81-203-0270-9

2. Linear Algebra and Its Applications by David C. Lay (4/e)


Publisher: Pearson Higher Education

ISBN: 0321385179

3. Linear Algebra (Third Edition) by Serge Lang


Publisher: Springer-Verlag New York, Inc.

ISBN: 0-387-96412-6

4. Matrix Theory and Linear Algebra by I. N. Herstein and D. J. Winter


Publisher: Macmillan Pub. Co.

ISBN: 978-0023539510



Hints/Solutions For Exercises

1. (a) 𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴1𝑦 = 𝑥1𝑦1 + 2𝑥1𝑦2 + 5𝑥1𝑦3 − 2𝑥2𝑦1 + 𝑥2𝑦3 − 6𝑥3𝑦2 + 6𝑥3𝑦3

(b) 𝑓 𝑥, 𝑦 = 𝑥𝑡𝐴2𝑦 = 3𝑥1𝑦1 + 𝑥1𝑦2 − 3𝑥1𝑦3 − 2𝑥2𝑦1 + 3𝑥2𝑦2

2. (a) 𝐴 =

2 0 3

0 2 5

4 0 0

(b) 𝐴 =

2 1 1

3 0 2

0 1 5

3. (a) 𝐴′ =0 0

0 4

(b) 𝐴′ =

1 1 11

5 2 11

5 1 1

4. (a) 𝑓 𝑥 = 𝑥𝑡𝐴3𝑥 = (𝑥2)2 + 2(𝑥3)2 + 10𝑥1𝑥2 − 2𝑥1𝑥3 + 12𝑥2𝑥3

(b) 𝑓 𝑥 = 𝑥𝑡𝐴4𝑥 = 𝑎𝑥2 + 𝑏𝑦2 + 𝑐𝑧2 + 2𝑕𝑥𝑦 + 2𝑓𝑦𝑧 + 2𝑔𝑧𝑥

5. (a) We have, by completing the squares,

2𝑦2 − 𝑧2 + 𝑥𝑦 + 𝑥𝑧 = 2(𝑦 +1

4𝑥)2 −

1

8(𝑥 − 4𝑧)2 + 𝑧2;

so that the matrix of the canonical form is 𝑃 =

2 0 0

10 0

8

0 0 1

.

Hence, rank is 3 and signature is 1.

(b) Setting 𝑥 = 𝑋 + 𝑌, 𝑦 = 𝑋 − 𝑌, 𝑧 = 𝑍, 𝑡 = 𝑇, we get

𝑦𝑧 + 𝑥𝑧 + 𝑥𝑦 + 𝑥𝑡 + 𝑦𝑡 + 𝑧𝑡 = (𝑋 + 𝑍 + 𝑇)2 − (𝑇 +1

2𝑍)2 −

3

4𝑍2 − 𝑌2;

so that the matrix of the canonical form is 𝑃 =

1 0 0 0

0 1 0 0

30 0 0

4

0 0 0 1

. Hence, rank is 4

and signature is 2.

lesson: bilinear, quadratic and hermitian forms lesson ... · 6. quadratic forms on ℝ 7....

Documents