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  Vector Spaces, Physics tutorial

Introduction to Vector Spaces:

Undoubtedly, you are quite well-known by the concept of a vector. By means of vector spaces, we are generalizing the fundamental idea. In another words, we shall encompass 'vectors' that are no longer just normal geometrical vectors, however vectors of a different type, however all encompassing similar properties. We shall come across the matrices that functions that you could give the similar treatment as you did in geometrical vectors.

Definition:

Given a set {v1, v2, ...., vn} = S. If

1) vi + vj ∈ S     ∀ i, j = 1, 2 ,...., n

2) α vi ∈ S     ∀i, = 1, 2 ,...., n;

α ∈ K, here 'K' is a field, example: the real number line (R) or the complex plane (C), then, 'S' is termed as a vector space or linear space. The vector space is a real vector space if K ≡ R and a complex vector space whenever K ≡ C.

The condition (1) states that whenever you add any two vectors of the vector space you will get a member of the space. Condition (2) exhibits that a linear multiplication of any two vectors generates a vector as well in the vector space.

Linear Independence:

Given a set {vi}i=1 n. If we can write:

a1v1 + a2v2 + ...... + anvn = 0

And this implies the constants a1 = a2 =..... = an = 0, then we state {vi}i=1 n is a linearly independent set.

If even just one of them is non-zero, then the set is linearly dependent. Think of it: a three-dimensional Cartesian vector will be a zero vector, 0, notice the boldface type (that is, not zero scalar), if and only if the three components are independently zero. Therefore, for instance, i, j, and k, the traditional unit vectors in three-dimensional Cartesian space are linearly independent. Mathematically, this signifies that:

αi + βj + γk = 0 if and only if α = β = γ = 0.

Basis Vector:

Assume that 'V' be an n-dimensional vector space. Any set of 'n' linearly independent vectors e1, e2, ...., en forms a basis for 'V'. Therefore, any vector v ∈ V can be deduced as a linear combination of the vectors e1, e2,...., en, that is,

X = x1e1 + x2e2 + .... xnen

Then we state that the vector space 'V' is spanned through the set of vectors {e1, e2, ...., en}.

{e1, e2, ...., en} is stated to be a basis for 'V'.

Whenever we wish for to write any vector in 1 (say, x) direction, we require only one (if possible, a unit) vector. Any two vectors in the x direction should be linearly dependent, for we can write one as a1i, and the other a2i, where a1 and a2 are scalars.

We make the linear combination:

c1 (a1i) + c2 (a2i) = 0

Here a1 and a2 are the scalar constants.

Evidently, c1 and c2 require not be zero for the expression to hold, for c1 = -c2 (a2/a1) would as well satisfy the expression.

We thus conclude that the vectors should be linearly dependent.

Inner or Scalar Product:

In this, we shall increase the idea of the inner product of two vectors.

Properties of the Inner Product:

Assume that 'V' be a vector space, real or complex. Then, the inner product of v, w ∈ V, written as (v, w), have the given properties:

1) (v,v) ≥ 0

2) (v,v) = 0 if and only if v = 0

3) (v,w) = (w,v)  (that is, Symmetry)

4) (cv,w) c * (v,w); (v, cw) = c(v,w)

5) (v, w + z) = (v,w) + (v,z)

6) (v, w) ≤ ||v||||w||

Here c* is the complex conjugate of the scalar 'c'.

Norm of a Vector:

Assume that 'X' be a vector space over 'K', then the real or complex number field. A real valued function ||.|| on X is a norm on X (that is, ||.||: X → R) if and only if the given conditions are satisfied:

1) ||x|| ≥ 0

2) ||x|| = 0 if and only if x = 0

3) || x + y || ≤ ||x|| + ||y|| ∀ x, y ∈ X (that is, Triangle inequality)

4) ||α x|| |α|||x|| ∀ x ∈ X and α ∈ C (that is, Absolute homogeneity)

The norm of a vector is the 'distance from the origin. Once again, you can observe the fundamental idea of the distance of a point from the origin being generalized to the case of the vectors in any vector space.

||x|| is known as the norm of 'x'.

In case where X = R, the real number line, the norm is the absolute value, |x|

When the norm of 'v' in the vector space 'V' is unity, such a vector is stated to be normalized. In any case, even when a vector is not normalized, we can normalize it by dividing by the norm.

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