MAT1341-L15-OrthogonalBasis0

MAT1341-L15-OrthogonalBasis0 - ORTHOGONAL AND ORTHONORMAL...

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Unformatted text preview: ORTHOGONAL AND ORTHONORMAL BASES ´ ´ ´ JOSE MALAGON-LOPEZ we have Among all the bases Rn has, the standard basis {e1, . . . , en } has the property that for any vector a1 . v=. . an v = a1 e1 + · · · + an en . An analogous easy description of a vector v in terms of a basis only happens for a special type of bases for Rn . It becomes clear what type of bases we are looking for once we realize that ei • v ei = ai ei . projei (v ) = ei 2 In other words, the coefficients describing v as a linear combination of {e1 , . . . , en } are of the form ei • v = ei • v, ei 2 and v = proje1 (v ) + · · · + projen (v ) . Notice that the properties of the standard unit vectors used here were that they were orthogonal among each other and that their norm was 1. These are the properties that a basis {v1, . . . , vn} should have if we want to obtain v = (vi • v ) v1 + · · · + (vi • v ) vn This lecture is about the description of such type of bases for Rn . Before we proceed, lets recall what is the norm of a vector and the projection of a vector along another vector. 1 The properties of the norm are The norm of a vector v in Rn is defined as v1 2 2 . v = . = v1 + · · · + vn . . vn (1) v ≥ 0, and v = 0 if and only if v = 0. (2) αv =| α | v , for any scalar α. (3) v 2 = v • v. Let v and u be two vectors in Rn , with u = 0. Then proju (v) is defined as v•u v•u proju (v) = u. u= u•u u2 Set u1 = proju(v ), and u2 = v − proju (v), so we have the following picture U1 V U U2 Remark 1. (1) If u and v are orthogonal, then v•u 0 proju (v ) = u= u = 0. u2 u2 (2) If αu = v , α = 0, then v•u u = αu proju (v) = u2 (3) Notice that the vectors u1 and u2 were defined so that u1 • u2 = 0 and v = u1 + u2 . 2 Orthogonal Basis Definition 2. Let B = {v1, . . . , vr } be a basis for a subspace V ⊆ Rn . We say that B is an orthogonal basis for V if for any 1 ≤ i = j ≤ r. Example 3. The set S = R2 . Indeed, this follows from −1 1 • 1 1 = (1)(−1) + (1)(1) = 0. vi • vj = 0, −1 1 , 1 1 is an orthogonal basis for 1 −1 1 1 , −1 , 1 is an orthogonal Example 4. The set S = 0 1 2 basis for R3 . Indeed, this follows from 1 1 1 • −1 = (1)(1) + (1)(−1) + (0)(1) = 0, 1 0 −1 1 1 • 1 = (1)(−1) + (1)(1) + (0)(2) = 0, 2 0 1 −1 1 • −1 = (−1)(1) + (1)(−1) + (2)(1) = 0. 1 2 Example 5. The set S = 1 1 • 1 2 1 1 , 1 2 2 for R . Indeed, this follows from is NOT an orthogonal basis = (1)(1) + (1)(2) = 3 = 0. 3 Before we continue, we state the following result. Theorem 6. Let S = {v1, . . . , vm} be a set of non-zero vectors such that vi • vj = 0, for any 1 ≤ i = j ≤ m. Then S is a linearly independent set. Proof. Let α1 , . . . , αm be scalars such that Then 0 = α1 v1 + · · · + αm vm. 0 = 0 • v1 = α1 (v1 • v1) + · · · + α1 (v1 • vm ) = α1 (v1 • v1) . Since v1 = 0 we have that v1 • v1 > 0, so we conclude that α1 = 0. We keep repeating this process with the remaining vectors in S so that we get α1 = α2 = · · · = αm = 0. Q.E.D. Theorem 7. Let S = {v1 , . . . , vr } be an orthogonal basis for a subspace V ⊆ Rn . Then any vector v in V can be written as v = projv1 (v ) + · · · + projvr (v). Proof. Since v is in V we have that there are scalars α1 , . . . , αr such that v = α1 v1 + · · · + αr vr . Then, for any 1 ≤ i ≤ r we have v • vi = (α1 v1 + · · · + αr vr ) • vi = αi (vi • vi) = αi vi 2 . Hence, αi = v • vi . Q.E.D. vi 2 Theorem 8. Any subspace V of Rn has an orthogonal basis. The proof of this theorem consists of constructing such orthogonal basis. The way of constructing such basis is called the Gram-Schmidt Orthogonalization Process. 4 Gram-Schmidt Orthogonalization Process Theorem 9 (Gram-Schmidt Orthogonalization Process). Assume that S = {v1 , . . . , vr } is a set of linearly independent vectors. Then the vectors u1 := v1 u2 := v2 − proju1 (v2) u3 := v3 − proju1 (v3) − proju2 (v3) . . . ur := vr − proju1 (vr ) − · · · − projur−1 (vr ) define an orthogonal basis for Span{v1, . . . , vr }. Proof. We will give the proof in steps. (1) u2 = 0 since S is a linearly independent set. (2) Since proju1 (v2) is collinear with u1 = v1, we have that Span{u1, u2} ⊆ Span{v1, v2}. (3) By construction, u1 • u2 = 0. (4) By the previous step and Theorem 6 we have that {u1, u2} is a linearly independent set. (5) By steps (2) and (4) we conclude that Span{u1, u2} = Span{v1, v2}. Now we repeat the previous steps. (6) u3 = 0. Indeed, if u3 = 0 then 0 = u3 = v3 − proju1 (v3) − proju2 (v3). This would imply that v3 is in Span{u1, u2} = Span{v1, v2}, which is not possible since S is a linearly independent set. 5 (7) Since u3 is a linear combination of u1, u2 and v3, we have that Span{u1, u2, u3} ⊆ Span{v1, v2, v3}. (8) u3 • u1 = 0 = u3 • u2. Indeed, u3 • u1 = v3 − proju1 (v3) − proju2 (v3) • u1 u1 • v3 u1 2 u1 • u1 − u2 • v3 u2 2 u2 • u1 = v3 • u1 − and = v3 • u1 − u1 • v3 = 0. u3 • u2 = v3 − proju1 (v3) − proju2 (v3) • u2 u1 • v3 u1 2 u1 • u2 − u2 • v3 u2 2 u2 • u2 = v3 • u2 − = v3 • u2 − 0 − u2 • v3 = 0. (9) By the previous step and Theorem 6 we have that {u1, u2, u3} is a linearly independent set. (10) By steps (7) and (9) we have that Span{u1 , u2, u3} = Span{v1, v2, v3}. (11) Repeating steps (6)-(10) for the next vectors until we run out of them will prove the theorem. Q.E.D. Example 10. Obtain an orthogonal basis for 2 2 1 1 , 0 , 2 . V = Span 1 1 0 First, notice that the vectors generating V are linearly independent. 6 Then, applying Gram-Schmidt process we obtain: 1 1 (1) u1 = 0 2 1 2 0 − (2) u2 = 1 0•1 1 1 1 0 −1 1= 2 1 0 1 1 0 1 2 1 2 2•1 2•−1 2 1 1 −1 0 1 1 1 1 1 − 2 −1 = 3 1 (3) u3 = 2 − 2 1 1 1 0 1 2 1 −1 0 1 Thus, 1 −1 1 1 , −1 , 1 1 3 1 2 0 is an orthogonal basis for V . is also an orthogonal basis for V . Remark 11. Notice that −1 1 1 1 , −1 , 1 2 1 0 7 Example 12. Using the result from the previous example, in this example we will illustrate the statement of theorem 7. 2 Let 2 be a vector in R3 . Since 2 −1 1 1 1 , −1 , 1 2 1 0 is an orthogonal basis for R3 , theorem 7 says that 2 2 2 = proj 2 1 2 2 1 0 2 proj 1 2 2 −1 1 + + 2 proj−1 2 2 1 2 1 2 −1 2 1 2 1 • 2 −1 • 2 1 • 2 1 −1 1 0 2 2 2 1 2 −1 + 1 1 + = 2 2 2 1 1 −1 1 2 0 1 −1 1 1 2 0 1 −1 1 2 2 −1 + 1 = 2 1 + 3 3 1 2 0 8 Orthonormal Basis Definition 13. An orthogonal basis B = {v1, . . . , vr } for a subspace V ⊆ Rn is called an orthonormal basis for V if vi = 1, for any 1 ≤ i ≤ r. Example 14. The set S = basis for R2 . Indeed, this follows from 11 √ 21 and 11 √ 21 1 −1 √ 21 1 =√ 2 1 =√ 2 • 1 −1 √ 21 = 1 0 ((1)(−1) + (1)(1)) = = 0. 2 2 √ 2 1 = √ = 1, 1 2 √ 2 −1 = √ =1 1 2 1 √ 2 1 √ −1 , 12 1 1 is an orthonormal −1 1 1 Example 15. The set S = 1 , −1 , 1 is an orthogonal 2 1 0 basis for R3 , but NOT an orthonormal basis for R3 . Indeed, we already saw that S is an orthogonal basis for R3 , but it is not orthonormal since 1 √ 1 = 2 = 1 0 Recall that given a vector v , we can always obtain a vector v ′ with the same direction that v but with norm 1: v v′ = . v Such vector v ′ is called the normalization of v . 9 Theorem 16. Any subspace V ⊆ Rn has an orthonormal basis. Proof. From theorem 8 the subspace V has an orthogonal basis {u1, . . . , ur }. Then the set of vectors u1 ur ,..., u1 ur is an orthonormal basis for V . Q.E.D. Example 17. We saw that 1 −1 1 S = 1 , −1 , 1 0 1 2 is an orthogonal basis for R3 . Then the orthonormal basis for R3 induced by S is 1 1 −1 −1 1 1 0 1 2 2 , 2 , 2 1 1 −1 −1 1 1 0 1 2 1 1 −1 1 1 1 −1 , √ 1 = √ 1 , √ 2 3 6 0 1 2 Example 18. Obtain an orthonormal basis for 1 1 1 1 , −2 , 0 V = Span −1 0 0 2 0 1 First notice that the three vectors generating V are linearly independent. 10 Next, we obtain an orthogonal basis: 1 1 (1) u1 = 0 1 1 1 −2 1 • 0 0 1 1 −2 1 0 1 (2) u2 = − 2 0 0 = 1 1 1 0 0 1 4 1 −5 3 0 1 14 1 1 0 −5 0 1 • • −1 0 −1 0 1 1 4 0 1 1 1 −5 2 2 (3) u3 = − 2 − 2 = −1 0 0 1 4 1 −5 1 2 1 0 0 1 1 Thus, 4 −4 1 1 1 1 −5 −2 S = , , 0 0 3 7 −1 1 1 6 −4 1 −2 7 −1 6 is an orthogonal basis for V . 11 Finally, we obtain an orthonormal basis by computing the normalizations of the vectors in S : 1 (1) 1 0 1 1 1 0 1 = 1 1 1 3 0, 1 4 −5 1 3 0 1 (2) 4 −5 1 3 0 1 −4 −2 1 7 −1 6 (3) −4 −2 1 7 −1 6 = 4 −5 1 3 0 1 4 −5 1 3 0 1 −4 −2 1 7 −1 6 −4 −2 1 7 −1 6 = 4 −5 0 1 4 −5 0 1 = 4 1 −5 42 0 1 = = −4 −2 −1 6 −4 −2 −1 6 = −4 1 −2 57 −1 6 E-mail address : jmalagon@uottawa.ca 1 4 −4 −5 1 −2 1 1 1 Thus, 3 , 42 , 57 is an orthonormal basis for V . 0 0 −1 1 1 6 12 ...
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