Book contents
- Frontmatter
- Dedication
- Contents
- Preface
- Notation and convention
- 1 Vector spaces
- 2 Linear mappings
- 3 Determinants
- 4 Scalar products
- 5 Real quadratic forms and self-adjoint mappings
- 6 Complex quadratic forms and self-adjoint mappings
- 7 Jordan decomposition
- 8 Selected topics
- 9 Excursion: Quantum mechanics in a nutshell
- Solutions to selected exercises
- Bibliographic notes
- References
- Index
4 - Scalar products
Published online by Cambridge University Press: 18 December 2014
- Frontmatter
- Dedication
- Contents
- Preface
- Notation and convention
- 1 Vector spaces
- 2 Linear mappings
- 3 Determinants
- 4 Scalar products
- 5 Real quadratic forms and self-adjoint mappings
- 6 Complex quadratic forms and self-adjoint mappings
- 7 Jordan decomposition
- 8 Selected topics
- 9 Excursion: Quantum mechanics in a nutshell
- Solutions to selected exercises
- Bibliographic notes
- References
- Index
Summary
In this chapter we consider vector spaces over a field which is either ℝ or ℂ. We shall start from the most general situation of scalar products. We then consider the situations when scalar products are non-degenerate and positive definite, respectively.
Scalar products and basic properties
In this section, we use F to denote the field ℝ or ℂ.
Definition 4.1 Let U be a vector space over F. A scalar product over U is defined to be a bilinear symmetric function f : U × U → F, written simply as (u, v) ≡ f(u, v), u, v ∈ U. In other words the following properties hold.
(Symmetry) (u, υ) = (υ, u) ∈ F for u, υ ∈ U.
(Additivity) (u + υ, w) = (u, w) + (υ, w) for u, υ, w ∈ U.
(Homogeneity) (au, υ) = a(u, υ) for a ∈ F and u, υ ∈ U.
We say that u, υ ∈ U are mutually perpendicular or orthogonal to each other, written as u ⊥ υ, if (u, υ) = 0. More generally for any non-empty subset S of U we use the notation
S⊥ = {u ∈ U | (u, υ) = 0 for any υ ∈ S}.
For u ∈ U we say that u is a null vector if (u, u) = 0.
It is obvious that S⊥ is a subspace of U for any nonempty subset S of U. Moreover {0}⊥ = U. Furthermore it is easy to show that if the vectors u1, …, uk are mutually perpendicular and not null then they are linearly independent.
Let u, υ ∈ U so that u is not null. Then we can resolve υ into the sum of two mutually perpendicular vectors, one in Span {u}, say cu for some scalar c, and one in Span{u}⊥, say w.
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- A Concise Text on Advanced Linear Algebra , pp. 115 - 146Publisher: Cambridge University PressPrint publication year: 2014