ORTHONORMAL BASIS: PROPERTIES, EXAMPLES AND EXERCISES - PHYSICAL - 2024

An orthonormal basis is formed with vectors perpendicular to each other and whose modulus is also 1 (unit vectors). Let us remember that a base B in a vector space V is defined as a set of linearly independent vectors capable of generating said space.

In turn, a vector space is an abstract mathematical entity among whose elements are vectors, generally associated with physical quantities such as speed, force and displacement or also with matrices, polynomials and functions.

Figure 1. Orthonormal base in the plane. Source: Wikimedia Commons. Quartl.

Vectors have three distinctive elements: magnitude or modulus, direction, and sense. An orthonormal basis is especially useful to represent and operate with them, since any vector that belongs to a certain vector space V can be written as a linear combination of the vectors that form the orthonormal basis.

In this way, operations between vectors, such as addition, subtraction and the different types of products defined in said space, are analytically executed.

Among the most commonly used bases in physics is the base formed by the unit vectors i, j, and k that represent the three distinctive directions of three-dimensional space: height, width, and depth. These vectors are also known as unit canonical vectors.

If, instead, the vectors are worked in a plane, two of these three components would suffice, while for one-dimensional vectors only one is required.

Properties of the bases

1- A base B is the smallest possible set of vectors that generate the vector space V.

2- The elements of B are linearly independent.

3- Any base B of a vector space V, allows to express all the vectors of V as a linear combination of it and this form is unique for each vector. That is why B is also known as the generating system.

4- The same vector space V can have different bases.

Examples of bases

Here are several examples of orthonormal bases and bases in general:

The canonical basis in ℜ

Also called natural base or standard base of ℜ ⁿ, where ℜ ⁿ is n-dimensional space, for example three-dimensional space is ℜ ³. The value of n is called the dimension of the vector space and is denoted as dim (V).

All vectors belonging to ℜ ⁿ are represented by ordered n-ads. For the space ℜ ⁿ, the canonical basis is:

e ₁ = <1,0,…, 0>; e ₂ = <0.1,…, 0>; …….. e _n = <0.0,…, 1>

In this example we have used the notation with brackets or “brackets” and bold for the unit vectors e ₁, e ₂, e ₃…

The canonical basis in ℜ

The familiar vectors i, j and k admit this same representation and all three of them are enough to represent the vectors in ℜ ³:

i = <1,0,0>; j = <0,1,0>; k = <0,0,1>

It means that the base can be expressed like this:

B = {<1,0,0>; <0,1,0>; <0,0,1>}

To verify that they are linearly independent, the determinant formed with them is non-zero and also equal to 1:

It must also be possible to write any vector belonging to ℜ ³ as a linear combination of them. For example, a force whose rectangular components are F _x = 4 N, F _y = -7 N and F _z = 0 N would be written in vector form like this:

F = <4, -7,0> N = 4 i -7 j + 0 k N.

Therefore i, j and k make up a generator system of ℜ ³.

Other orthonormal bases in ℜ

The standard base described in the previous section is not the only orthonormal base in ℜ ³. Here we have for example the bases:

B ₁ = {; <- sin θ, cos θ, 0>; <0,0,1>}

B ₂ = {<3/5, 4 / 5.0>; <- 4/5, 3 / 5.0>; <0,0,1>}

It can be shown that these bases are orthonormal, for this we remember the conditions that must be met:

-The vectors that form the base must be orthogonal to each other.

-Each of them must be unitary.

We can verify this by knowing that the determinant formed by them must be non-zero and equal to 1.

The base B ₁ is precisely that of cylindrical coordinates ρ, φ and z, another way of expressing vectors in space.

Figure 2. Cylindrical coordinates. Source: Wikimedia Commons. Math buff.

Solved exercises

- Exercise 1

Show that the base B = {<3/5, 4 / 5,0>; <- 4/5, 3 / 5.0>; <0,0,1>} is orthonormal.

Solution

To show that the vectors are perpendicular to each other, we will use the scalar product, also called the internal or dot product of two vectors.

Let any two vectors u and v, their dot product is defined by:

u • v = uv cosθ

To distinguish the vectors of their modules we will use bold for the first and normal letters for the second. θ is the angle between u and v, therefore if they are perpendicular, it means that θ = 90º and the scalar product is zero.

Alternatively, if the vectors are given in terms of their components: u =x, u _y, u _z > y v =x, v _y, v _z >, the scalar product of both, which is commutative, is calculated in this way:

u • v = u _x.v _x + u _y.v _y + u _z.v _z

In this way, the scalar products between each pair of vectors are, respectively:

i) <3/5, 4 / 5,0> • <- 4/5, 3 / 5,0> = (3/5). (- 4/5) + (4/5). ((3 / 5) + 0.0 = (-12/25) + (12/25) = 0

ii) <3/5, 4 / 5.0> • <0, 0.1> = 0

iii) <- 4/5, 3 / 5.0> • <0, 0.1> = 0

For the second condition, the module of each vector is calculated, which is obtained by:

│u │ = √ (u _x ² + u _y ² + u _z ²)

Thus, the modules of each vector are:

│ <3/5, 4 / 5,0> │ = √ = √ = √ (25/25) = 1

│ <-4/5, 3 / 5,0> │ = √ = √ = √ (25/25) = 1

│ <0, 0.1> │ = √ = 1

Therefore all three are unit vectors. Finally, the determinant they form is non-zero and equal to 1:

- Exercise 2

Write the coordinates of the vector w = <2, 3,1> in terms of the base above.

Solution

To do this, the following theorem is used:

w = < w • v ₁ > v ₁ + < w • v ₂ > v ₂ + < w • v ₃ > v ₃ +… < w • v _n > v _n

This means that we can write the vector in base B, using the coefficients < w • v ₁ >, < w • v ₂ >,… < w • v _n >, for which we must calculate the indicated scalar products:

<2, 3,1> • <3/5, 4 / 5,0> = (2). (3/5) + (3). (4/5) + 1.0 = (6/5) + (12 / 5) = 18/5

<2, 3,1> • <- 4/5, 3 / 5,0> = (2). (- 4/5) + (3). (3/5) + 1.0 = (-8/5) + (9/5) = 1/5

<2, 3,1> • <0,0,1> = 1

With the scalar products obtained, a matrix is ​​constructed, called the w coordinate matrix.

Therefore the coordinates of the vector w in the base B are expressed by:

_B =

The coordinate matrix is ​​not the vector, since a vector is not the same as its coordinates. These are only a set of numbers that serve to express the vector in a given base, not the vector as such. They also depend on the selected base.

Finally, following the theorem, the vector w would be expressed as follows:

w = (18/5) v ₁ + (1/5) v ₂ + v ₃

With: v ₁ = <3/5, 4 / 5,0>; v ₂ = <- 4/5, 3 / 5.0>; v ₃ = <0,0,1>}, that is, the vectors of the base B.

References

1. Larson, R. Foundations of Linear Algebra. 6th. Edition. Cengage Learning.
2. Larson, R. 2006. Calculus. 7th. Edition. Volume 2. McGraw Hill.
3. Salas, J. Linear Algebra. Unit 10. Orthonormal bases. Recovered from: ocw.uc3m.es.
4. Sevilla University. Cylindrical coordinates. Vector base. Recovered from: laplace.us.es.
5. Wikipedia. Orthonormal base. Recovered from: es.wikipedia.org.