*Notes to a video lecture on http://www.unizor.com*

__Vectors+ 08 - Hilbert Space__

Let's continue building our abstract theory of

**vector spaces**introduced in the previous lecture

*Vectors 07*of this

*Vectors*chapter of this course

*Math+ & Problems*on

*UNIZOR.COM*.

Our next addition is a

**scalar product**of two elements of an abstract vector space.

In case of

*N*-dimensional Euclidian space with two vectors

*and*

**R**(R_{1},R_{2},...,R_{N})

**S**(S_{1},S_{2},...,S_{N})we

__defined__scalar product as

**R·S**= R_{1}·S_{1}+R_{2}·S_{2}+...+R_{N}·S_{N}Thus defined, the scalar product had certain properties and characteristics that we have proven based on this definition.

In case of abstract vectors spaces, the scalar product

__is not explicitly defined__, but, instead, defined as

__any function of two vectors from our vector space that satisfies certain axioms__that resemble the properties and characteristics of a scalar product of two vectors in

*N*-dimensional Euclidian space.

Let's assume, we have a vector space

*and a scalar space*

**V***associated with it - a set of all real numbers in our case.*

**S**All axioms needed for

*and*

**V***to be a vector space were described in the previous lecture mentioned above.*

**S**Now we assume that for any two vectors

*and*

**a***from*

**b***there exists a real number called their*

**V****scalar product**denoted

*that satisfies the following set of axioms.*

**a·b**(1) For any vector

*from*

**a***, which is not a*

**V****null-vector**, its scalar product with itself is a positive real number

∀

*∈*

**a***,*

**V***:*

**a≠0**

**a·a >**0(2) For null-vector

*its scalar product with itself is equal to zero*

**0***:*

**a=0**

**a·a =**0(3) Scalar product of any two vectors

*and*

**a***is*

**b****commutative**

∀

*∈*

**a,b***:*

**V**

**a·b = b·a**(4) Scalar product of any two vectors

*and*

**a***is proportional to their magnitude*

**b**∀

*∈*

**a,b***, ∀*

**V***γ*∈

*:*

**S**

**(**γ**·a)·b =**γ**·(a·b)**(5) Scalar product is

**distributive**relatively to addition of vectors. ∀

*∈*

**a,b,c***:*

**V**

**(a+b)·c = a·c+b·c**As before, a square root from a scalar product of a vector by itself will be called

**magnitude**or

**length**, or

**norm**of this vector

**||a|| = √(a·a)**Using the above defined scalar product, we can define a

**distance**between vectors

*and*

**a***as an absolute value of a magnitude of vector*

**b***, which we will write as*

**a+(−b)***.*

**a−b**

**||a−b|| = √(a−b)·(a−b)**A vector space with a scalar product defined above is called a

**pre-Hilbert space**.

To be a

**Hilbert space**, we need one more condition - the

**completeness**of vector space, which means that every converging in a certain way sequence of vectors has a vector that is the limit of this sequence within the same vector space.

More rigorously, the convergence is defined in terms of

**Cauchy criterion**. It states that a sequence of vectors {

*} converges if*

**a**_{i}for any

*ε>0*there is a natural number

*N*such that the distance between

*and*

**a**_{m}*is less than*

**a**_{n}*ε*for any

*m,n ≥ N*

∀

*ε>0*∃

*N: m,n > N*=> ||

*||<*

**a**_{m}−a_{n}*ε*

*Problem A*

Prove that a scalar product of null-vector with any other is zero.

*Proof A*

∀

*∈*

**a***:*

**V**

**0·a = (**0**·a)·a =**0**·(a·a) =**0End of proof.

*Problem B*

Prove that a scalar product changes the sign, if one of its components is replaced with its opposite.

*Proof B*

[

*Problem C*from the previous lecture

*Vectors 07*stated that

*]*

**−a=**−1**·a**∀

*∈*

**a,b***:*

**V**

**(−a)·b = (**−1**·a)·b =**−1**·(a·b)**End of proof.

*Problem C*

Prove the Cauchy-Schwartz-Bunyakovsky inequality

∀

*∈*

**a,b***:*

**V***.*

**(a·b)² ≤ (a·a)·(b·b)***Proof C*

If either

*or*

**a***equals to null-vector, we, obviously, get zero on both sides of inequality, which satisfies the sign*

**b****≤**.

Assume, both vectors are not-null.

Consider any non-zero

*γ*and non-negative scalar product of

*by itself*

**a+**γ**b***0 ≤*

**(a+**γ**·b)·(a+**γ**·b)**Use commutative and distributive properties to open all parenthesis

*0 ≤*

**a·a+**2γ**·a·b+**γ²**·b·b**Set

*γ*[

**=−(a·b)/***]*

**(b·b)**With this the inequality above takes form

*0 ≤*

**a·a−**2**·(a·b)²/(b·b)+(a·b)²/(b·b)**Multiplying this inequality by a positive

*, we obtain*

**b·b***0 ≤*

**(a·a)·(b·b)−**2**·(a·b)²+(a·b)²**which transforms into

**(a·b)² ≤ (a·a)·(b·b)**End of proof.