Orthonormal basis properties, examples and exercises

A orthonormal base It is formed with vectors perpendicular to each other and whose module is also equal to 1 (unit vectors). Let's 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..

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 are analytically executed, such as addition, subtraction and the different types of products defined in said space..

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

Article index

• 1 Properties of the bases
• 2 Examples of bases
  • 2.1 The canonical basis in ℜ n
  • 2.2 The canonical basis on ℜ3
  • 2.3 Other orthonormal bases in ℜ3
• 3 Solved exercises
  • 3.1 - Exercise 1
  • 3.2 - Exercise 2
• 4 References

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 they are linearly independent.

3- Any base B of a vector space V, allows to express all vectors of V as a linear combination of it and this shape is unique for each vector. Therefore to B it is also known as generator 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 ⁿ

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

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

and₁ = <1,0,… ,0>; and_two = <0,1,… ,0>; ... and_n = <0,0,… ,1>

In this example we have used the notation with brackets or “brackets” and bold for the unit vectors and₁, and_two, and₃...

The canonical basis in ℜ³

The familiar vectors i, j Y 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 should also be possible to write any vector that belongs 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 = 4i -7j + 0k N.

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

Other orthonormal bases in ℜ³

The standard base described in the previous section is not the only orthonormal base in ℜ³. Here we have for example the bases:

B₁ = ; <- sen θ, cos θ,0 >;  <0,0,1 >

B_two = <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 it knowing that the determinant formed by them must be non-zero and equal to 1.

Base B₁ it is precisely that of cylindrical coordinates ρ, φ and z, another way of expressing vectors in space.

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 be any two vectors or Y v, its dot product is defined by:

or • v = u.v. cosθ

To distinguish the vectors of their modules we will use bold for the first ones and normal letters for the second ones. θ is the angle between or Y 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: or = x, or_Y,or_z > and v = x, v_Y,v_z >, the dot product of both, which is commutative, is calculated as follows:

or • v = or_x .v_x + or_Y .v_Y + or_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^two + or_Y^two + or_z^two)

Thus, the modules of each vector are:

│<3/5, 4/5,0 >│ = √ [(3/5)^two + (4/5)^two  + 0^two)] = √ [(9/25) + (16/25)] = √ (25/25) = 1

│<-4/5, 3/5,0 >│ = √ [(-4/5)^two + (3/5)^two  + 0^two)] = √ [(16/25) + (9/25)] = √ (25/25) = 1

│<0, 0,1 >│ = √ [0^two + 0^two  + 1^two)] = 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 old base.

Solution

To do this, the following theorem is used:

Let B = v₁, v_two, v₃,... v_n an orthonormal basis in space V with inner product, the vector w is represented by B as follows:

w = <w•v₁> v₁ + <w•v_two> v_two +<w•v₃> v₃ +... <w•v_n> v_n

This means that we can write the vector in base B, using the coefficients <w•v₁>, <w•v_two>, ...  <w•v_n>, for which the indicated scalar products must be calculated:

<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 ​​built, called coordinate matrix of w.

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

[w]_B= [(5/18); (1/5); 1]

The coordinate matrix is ​​not the vector, since a vector It 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 it would be expressed like this:

w = (18/5) v₁ + (1/5) v_two + v₃

With: v₁ = <3/5, 4/5,0 >; v_two = <- 4/5, 3/5,0 >; v₃ =  <0,0,1 >, that is, the vectors of the basis 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.