INTRODUCTION

In Section 7.6 we saw that a vector space V can have many different bases. Recall, the defining characteristics of any basis B = {x₁, x₂, … , x_n} of a vector space V is that

• the set B is linearly independent, and
• the set B spans the space.

In this context the word span means that every vector in the space can be expressed as a linear combination of the vectors x₁, x₂, … , x_n. For example, every vector u in Rⁿ can be written as a linear combination of the vectors in the standard basis B = {e₁, e₂, … , e_n}, where

e₁ = 〈1, 0, 0, … , 0〉, e₂ = 〈0, 1, 0, … , 0〉, …, e_n = 〈0, 0, 0, … , 1〉.

This standard basis B = {e₁, e₂, … , e_n} is also an example of an orthonormal basis; that is, the e_i, i = 1, 2, … , n are mutually orthogonal and are unit vectors; that is,

e_i · e_j = 0, i ≠ j and e_i = 1, i = 1, 2, … , n.

In this section we focus on orthonormal bases for Rⁿ and examine a procedure whereby we can transform or convert any basis B of Rⁿ into an orthonormal basis.

EXAMPLE 1 Orthonormal Basis for R³

A basis B for Rⁿ need not be orthogonal nor do the basis vectors need to be unit vectors. In fact, any linearly independent set of n vectors can serve as a basis for the n-dimensional vector space Rⁿ. For example, it is a straightforward task to show that the vectors

u₁ = 〈1, 0, 0〉, u₂ = 〈1, 1, 0〉, u₃ = 〈1, 1, 1〉

in R³ are linearly independent, and hence B = {u₁, u₂, u₃} is a basis for R³. Note that B is not an orthogonal basis.

Generally, an orthonormal basis for a vector space V turns out to be the most convenient basis for V. One of the advantages that an orthonormal basis has over any other basis for Rⁿ is the comparative ease with which we can obtain the coordinates of a vector u relative to that basis.

PROOF:

The vector u is in Rⁿ and so it is an element of the set Span(B). In other words, there exist real scalars k_i, i = 1, 2, … , n such that u can be expressed as the linear combination

u = k₁w₁ + k₂w₂ + … + k_nw_n.

The scalars k_i are the coordinates of u relative to the basis B. These coordinates can be found by taking the dot product of u with each of the basis vectors:

u · w_i = (k₁w₁ + k₂w₂ + … + k_nw_n) · w_i = k₁(w₁ · w_i) + k₂(w₂ · w_i) + … + k_n(w_n · w_i). (2)

Since B is orthonormal, w_i is orthogonal to all vectors in B with the exception of w_i itself. That is, w_i · w_j = 0, i ≠ j and w_i · w_i = w_i² = 1. Hence from (2), we obtain k_i = (u · w_i) for i = 1, 2, … , n. ≡

EXAMPLE 2 Coordinates of a Vector in R³

Find the coordinates of the vector u = 〈3, −2, 9〉 relative to the orthonormal basis B for R³ given in (1) of Example 1. Write u in terms of the basis B.

SOLUTION

From Theorem 7.7.1, the coordinates of u relative to the basis B in (1) of Example 1 are simply

Hence we can write

≡

Gram–Schmidt Orthogonalization Process

The procedure known as the Gram–Schmidt orthogonalization process is a straightforward algorithm for generating an orthogonal basis B′ = {v₁, v₂, … , v_n} from any given basis B = {u₁, u₂, … , u_n} for Rⁿ. We then produce an orthonormal basis B″ = {w₁, w₂, … , w_n} by normalizing the vectors in the orthogonal basis B′. The key idea in the orthogonalization process is vector projection, and so we suggest that you review that concept in Section 7.3. Also, for the sake of gaining some geometric insight into the process, we shall begin in R² and R³.

Constructing an Orthogonal Basis for R²

The Gram–Schmidt orthogonalization process for Rⁿ is a sequence of steps; at each step we construct a vector v_i that is orthogonal to the vector in the preceding step. The transformation of a basis B = {u₁, u₂} for R² into an orthogonal basis B′ = {v₁, v₂} consists of two steps. See FIGURE 7.7.1(a). The first step is simple, we merely choose one of the vectors in B, say, u₁, and rename it v₁. Next, as shown in Figure 7.7.1(b), we project the remaining vector u₂ in B onto the vector v₁ and define a second vector to be v₂ = u₂ − u₂. Recall from (12) of Section 7.3 that u₂ = . As seen in Figure 7.7.1(c), the vectors

(3)

are orthogonal. If you are not convinced of this, we suggest you verify the orthogonality of v₁ and v₂ by demonstrating that v₁ · v₂ = 0.

EXAMPLE 3 Gram–Schmidt Process in R²

The set B = {u₁, u₂}, where u₁ = 〈3, 1〉, u₂= 〈1, 1〉, is a basis for R². Transform B into an orthonormal basis B″ = {w₁, w₂}.

SOLUTION

We choose v₁ as u₁: v₁ = 〈3, 1〉. Then from the second equation in (3), with u₂ · v₁ = 4 and v₁ · v₁ = 10, we obtain

The set B′ = {v₁, v₂} = {〈3, 1〉, 〈−, 〉} is an orthogonal basis for R². We finish by normalizing the vectors v₁ and v₂:

.

The basis B is shown in FIGURE 7.7.2(a), and the new orthonormal basis B″ = {w₁, w₂} is shown in blue in Figure 7.7.2(b). ≡

In Example 3 we are free to choose either vector in B = {u₁, u₂} as the vector v₁. However, by choosing v₁ = u₂ = 〈1, 1〉, we obtain a different orthonormal basis, namely, B″ = {w₁, w₂}, where w₁ = 〈1/, 1/〉 and w₂ = 〈1/, −1/〉. See Problems 5–8 in Exercises 7.7.

Constructing an Orthogonal Basis for R³

Now suppose B = {u₁, u₂, u₃} is a basis for R³. Then the set B′ = {v₁, v₂, v₃}, where

(4)

is an orthogonal basis for R³. Again, if you do not see this, then compute v₁· v₂, v₁ · v₃, and v₂ · v₃.

Since the vectors v₁ and v₂ in the list (4) are by construction orthogonal, the set {v₁, v₂} must be linearly independent. See Problem 36 in Exercises 7.6. Thus, W₂ = Span(v₁, v₂) is necessarily a two-dimensional subspace of R³. Now the vector is a vector in W₂ because it is a linear combination of v₁ and v₂. The vector x is called the orthogonal projection of u₃ onto the subspace W₂ and is usually denoted by In FIGURE 7.7.3, x is the red vector. Notice, too, that x is the sum of two projections. Using (12) of Section 7.3, we can write

.(5)

The difference v₃ = u₃ − x is orthogonal to x. Indeed, v₃ is orthogonal to v₁ and v₂ and to every vector in W₂. This is precisely the same idea in (3). In that context, v₂ = u₂ − x, where x was the projection of u₂ onto the one-dimensional subspace W₁ = Span(v₁) of R². Analogous to (5), we have

(6)

EXAMPLE 4 Gram–Schmidt Process in R³

We conclude this section with a theorem that summarizes the most general case of the Gram–Schmidt process for Rⁿ. The orthogonalization process can be used on any linearly independent set S, and so we can use it to find orthonormal bases for subspaces of Rⁿ.

7.7 Exercises Answers to selected odd-numbered problems begin on page ANS-17.

In Problems 1 and 2, verify that the basis B for the given vector space is orthonormal. Use Theorem 7.7.1 to find the coordinates of the vector u relative to the basis B. Then write u as a linear combination of the basis vectors.

In Problems 3 and 4, verify that the basis B for the given vector space is orthogonal. Use Theorem 7.7.1 as an aid in finding the coordinates of the vector u relative to the basis B. Then write u as a linear combination of the basis vectors.

In Problems 5−8, use the Gram–Schmidt orthogonalization process (3) to transform the given basis B = {u₁, u₂} for R² into an orthogonal basis B′ = {v₁, v₂}. Then form an orthonormal basis B″ = {w₁, w₂}.

1. First construct B″ using v₁, u₁.
2. Then construct B″ using v₁, u₂.
3. Sketch B and each basis B″.

5. B = {〈−3, 2〉, 〈−1, −1〉}
6. B = {〈−3, 4〉, 〈−1, 0〉}
7. B = {〈1, 1〉, 〈1, 0〉}
8. B = {〈5, 7〉, 〈1, −2〉}

In Problems 9–12, use the Gram–Schmidt orthogonalization process (4) to transform the given basis B = {u₁, u₂, u₃} for R³ into an orthogonal basis B′ = {v₁, v₂, v₃}. Then form an orthonormal basis B″ = {w₁, w₂, w₃}.

9. B = {〈1, 1, 0〉, 〈1, 2, 2〉, 〈2, 2, 1〉}
10. B = {〈−3, 1, 1〉, 〈1, 1, 0〉, 〈−1, 4, 1〉}
11. B = {〈, , 1〉, 〈−1, 1, −〉, 〈−1, , 1〉}
12. B = {〈1, 1, 1〉, 〈9, −1, 1〉, 〈−1, 4, −2〉}

In Problems 13 and 14, the given vectors span a subspace W of R³. Use the Gram–Schmidt orthogonalization process to construct an orthonormal basis for the subspace.

13. u₁ = 〈1, 5, 2〉, u₂ = 〈−2, 1, 1〉
14. u₁ = 〈1, 2, 3〉, u₂ = 〈3, 4, 1〉

In Problems 15 and 16, the given vectors span a subspace W of R⁴. Use the Gram–Schmidt orthogonalization process to construct an orthonormal basis for the subspace.

15. u₁ = 〈1, −1, 1, −1〉, u₂ = 〈1, 3, 0, 1〉
16. u₁ = 〈4, 0, 2, −1〉, u₂ = 〈2, 1, −1, 1〉, u₃ = 〈1, 1, −1, 0〉

In Problems 17 and 18, an inner product defined on the vector space P₂ of all polynomials of degree less than or equal to 2, is given by

Use the Gram–Schmidt orthogonalization process to transform the given basis B for P₂ into an orthogonal basis B′.

17. B = {1, x, x²}
18. B = {x² − x, x² + 1, 1 − x²}

For the inner product (p, q) defined on P₂ in Problems 17 and 18, the norm of a polynomial p is defined by

Use this norm in Problems 19 and 20.

19. Construct an orthonormal basis B″ from B′ obtained in Problem 17.
20. Construct an orthonormal basis B″ from B′ obtained in Problem 18.

In Problems 21 and 22, let p(x) = 9x² − 6x + 5 be a vector in P₂. Use Theorem 7.7.1 and the indicated orthonormal basis B″ to find the coordinates p(x) relative to B″. Then write p(x) as a linear combination of the basis vectors.

21. B″ in Problem 19
22. B″ in Problem 20

Discussion Problem

23. The set of vectors {u₁, u₂, u₃}, where
    

    u₁ = 〈1, 1, 3〉, u₂ = 〈1, 4, 1〉, and u₃ = 〈1, 10, −3〉,

    is linearly dependent in R³ since u₃ = −2u₁ + 3u₂. Discuss what you would expect when the Gram–Schmidt process in (4) is applied to these vectors. Then carry out the orthogonalization process.