## Saturday, August 10, 2024

### Vectors+ 08 Hilbert Space: UNIZOR.COM - Math+ & Problems - Vectors

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
R(R1,R2,...,RN) and
S(S1,S2,...,SN)
we defined scalar product as
R·S = R1·S1+R2·S2+...+RN·SN

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 V and a scalar space S associated with it - a set of all real numbers in our case.
All axioms needed for V and S to be a vector space were described in the previous lecture mentioned above.
Now we assume that for any two vectors a and b from V there exists a real number called their scalar product denoted a·b that satisfies the following set of axioms.

(1) For any vector a from V, which is not a null-vector, its scalar product with itself is a positive real number
aV, a≠0: a·a > 0
(2) For null-vector 0 its scalar product with itself is equal to zero
a=0: a·a = 0
(3) Scalar product of any two vectors a and b is commutative
a,bV: a·b = b·a
(4) Scalar product of any two vectors a and b is proportional to their magnitude
a,bV, ∀γS: (γ·a)·b = γ·(a·b)
(5) Scalar product is distributive relatively to addition of vectors. ∀a,b,cV: (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 a and b as an absolute value of a magnitude of vector a+(−b), which we will write as 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 {ai} converges if
for any ε>0 there is a natural number N such that the distance between am and an is less than ε for any m,n ≥ N
ε>0N: m,n > N => ||am−an||<ε

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

Proof A
aV: 0·a = (0·a)·a = 0·(a·a) = 0
End 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,bV: (−a)·b = (−1·a)·b = −1·(a·b)
End of proof.

Problem C
Prove the Cauchy-Schwartz-Bunyakovsky inequality
a,bV: (a·b)² ≤ (a·a)·(b·b).

Proof C
If either a or b equals to null-vector, we, obviously, get zero on both sides of inequality, which satisfies the sign .
Assume, both vectors are not-null.
Consider any non-zero γ and non-negative scalar product of a+γb by itself
0 ≤ (a+γ·b)·(a+γ·b)
Use commutative and distributive properties to open all parenthesis
0 ≤ a·a+·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 b·b, we obtain
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.