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SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 455 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS All the vector spaces we have studied thus far in the text are real vector spaces since the scalars are real numbers. A complex vector space is one in which the scalars are complex numbers. Thus, if v1, v2, . , vm are vectors in a complex vector space, then a linear com- bination is of the form 1 1 ? ? ? 1 c1v1 c 2v2 cmvm n where the scalarsc1, c 2, . , cm are complex numbers. The complex version of R is the complex vector spaceC n consisting of ordered n-tuples of complex numbers. Thus, a vector in C n has the form 5 s 1 1 1 d v a1 b1i, a 2 b2i, . , an bni . It is also convenient to represent vectors in C n by column matrices of the form 1 a1 b1i a 1 b i v 5 2 . 2 . 3 . 4 1 an bni As with Rn, the operations of addition and scalar multiplication in C n are performed com- ponent by component. EXAMPLE 1 Vector Operations in Cn Let v 5 s1 1 2i, 3 2 id and u 5 s22 1 i, 4d be vectors in the complex vector spaceC 2 , and determine the following vectors. (a)v 1 u (b)s2 1 idv (c) 3v 2 s5 2 idu Solution (a) In column matrix form, the sum v 1 u is 1 1 2i 22 1 i 21 1 3i v 1 u 5 3 4 1 3 4 5 3 4. 3 2 i 4 7 2 i (b) Since s2 1 ids1 1 2id 5 5i and s2 1 ids3 2 id 5 7 1 i, we have s2 1 idv 5 s2 1 ids1 1 2i, 3 2 id 5 s5i, 7 1 id. (c) 3v 2 s5 2 idu 5 3s1 1 2i, 3 2 id 2 s5 2 ids22 1 i, 4d (c) 5 s3 1 6i, 9 2 3id 2 s29 1 7i, 20 2 4id (c) 5 s12 2 i, 211 1 id 456 CHAPTER 8 COMPLEX VECTOR SPACES Many of the properties ofRn are shared by Cn. For instance, the scalar multiplicative identity is the scalar 1 and the additive identity in Cn is 0 5 s0, 0, 0, . , 0d. The standard basis for Cn is simply 5 s d e1 1, 0, 0, . , 0 5 s d e2 0, 1, 0, . , 0 . 5 s d en 0, 0, 0, . , 1 which is the standard basis for Rn. Since this basis contains n vectors, it follows that the dimension of Cn is n. Other bases exist; in fact, any linearly independent set of n vectors in Cn can be used, as we demonstrate in Example 2. EXAMPLE 2 Verifying a Basis Show that v v v { 1 { 2 { 3 S 5 hsi, 0, 0d, si, i, 0d, s0, 0, idj is a basis for C 3. 3 h j Solution Since C has a dimension of 3, the set v1, v2, v3 will be a basis if it is linearly indepen- dent. To check for linear independence, we set a linear combination of the vectors in S equal to 0 as follows. 1 1 5 s d c 1v1 c2v2 c3v3 0, 0, 0 s d 1 s d 1 s d 5 s d c1i, 0, 0 c2i, c2i, 0 0, 0, c3i 0, 0, 0 ss 1 d d 5 s d c1 c2 i, c2i, c3i 0, 0, 0 This implies that s 1 d 5 c1 c2 i 0 5 c 2i 0 5 c 3i 0. 5 5 5 h j Therefore,c1 c2 c3 0, and we conclude thatv1, v2, v3 is linearly independent. EXAMPLE 3 Representing a Vector in Cn by a Basis Use the basis S in Example 2 to represent the vector v 5 s2, i, 2 2 id. SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 457 Solution By writing 5 1 1 v c1v1 c2v2 c3v3 5 ss 1 d d c1 c2 i, c2i, c3i 5 s2, i, 2 2 id we obtain s 1 d 5 c1 c2 i 2 5 c 2i i 5 2 c 3i 2 i 5 which implies that c2 1 and 2 2 i 2 2 i c 5 521 2 2i and c 5 521 2 2i. 1 i 3 i Therefore, 5 s2 2 d 1 1 s2 2 d v 1 2i v1 v2 1 2i v3. Try verifying that this linear combination yields s2, i, 2 2 id. Other than Cn, there are several additional examples of complex vector spaces. For instance, the set of m 3 n complex matrices with matrix addition and scalar multiplication forms a complex vector space. Example 4 describes a complex vector space in which the vectors are functions. EXAMPLE 4 The Space of Complex-Valued Functions Consider the set S of complex-valued functions of the form s 5 s d 1 s d f x) f1 x if 2 x wheref1 and f2 are real-valued functions of a real variable. The set of complex numbers form the scalars for S and vector addition is defined by s d 1 s d 5 f s d 1 s dg 1 f d 1 s dg f x g x f1 x if 2 x g1(x i g2 x 5 f s d 1 s dg 1 f s d 1 s dg f1 x g1 x i f 2 x g2 x . It can be shown that S, scalar multiplication, and vector addition form a complex vector space. For instance, to show that S is closed under scalar multiplication, we let c 5 a 1 bi be a complex number. Then s d 5 s 1 df s d 1 s dg c f x a bi f1 x if 2 x 5 f s d 2 s dg 1 f s d 1 s dg af1 x bf 2 x i bf1 x af 2 x is in S. 458 CHAPTER 8 COMPLEX VECTOR SPACES The definition of the Euclidean inner product in Cn is similar to that of the standard dot product in Rn, except that here the second factor in each term is a complex conjugate. Definition of Let u and v be vectors in Cn. The Euclidean inner product of u and v is given by Euclidean Inner ? 5 1 1 ? ? ? 1 u v u1v1 u2v2 unvn. Product in C n REMARK: Note that if u and v happen to be “real,” then this definition agrees with the standard inner (or dot) product in R n. EXAMPLE 5 Finding the Euclidean Inner Product in C3 Determine the Euclidean inner product of the vectors u 5 s2 1 i, 0, 4 2 5id and v 5 s1 1 i, 2 1 i, 0d. Solution ? 5 1 1 u v u1v1 u2v2 u 3v3 5 s2 1 ids1 2 id 1 0s2 2 id 1 s4 2 5ids0d 5 3 2 i Several properties of the Euclidean inner productCn are stated in the following theorem. Theorem 8.7 Let u, v, and w be vectors in C n and let k be a complex number. Then the following Properties of the properties are true. 1. u ? v 5 v ? u Euclidean Inner 2. su 1 vd ? w 5 u ? w 1 v ? w Product 3. skud ? v 5 ksu ? vd 4. u ? skvd 5 ksu ? vd 5. u ? u $ 0 6. u ? u 5 0if and only if u 5 0. Proof We prove the first property and leave the proofs of the remaining properties to you. Let 5 s d 5 s d u u1, u 2, . , un and v v1, v2, . , vn . Then ? 5 1 1 . 1 v u v1u1 v2u2 vnun 5 1 1 . 1 v1u1 v2u 2 vnun 5 1 1 . 1 v1u1 v2u 2 vnun SECTION 8.4 COMPLEX VECTOR SPACES AND INNER PRODUCTS 459 5 1 1 ? ? ? 1 u1v1 u2v2 unvn 5 u? v. We now use the Euclidean inner product in Cn to define the Euclidean norm (or length) of a vector in Cn and the Euclidean distance between two vectors in Cn. Definition of Norm The Euclidean norm (or length) of u in Cn is denoted by iui and is n and Distance in C iui 5 su ? ud1y2. The Euclidean distance between u and v is dsu, vd 5 iu 2 vi . The Euclidean norm and distance may be expressed in terms of components as i i 5 su u 2 1 u u 2 1 ? ? ? 1 u u 2d1y2 u u1 u2 un s d 5 su 2 u 2 1 u 2 u 2 1 ? ? ? 1 u 2 u 2d1y2 d u, v u1 v1 u2 v2 un vn . EXAMPLE 6 Finding the Euclidean Norm and Distance in C n Determine the norms of the vectors u 5 s2 1 i, 0, 4 2 5id and v 5 s1 1 i, 2 1 i, 0d and find the distance between u and v. Solution The norms of u and v are given as follows. i i 5 su u 2 1 u u 2 1 u u 2d1y2 u u1 u2 u3 2 2 2 2 2 2 1y2 5 fs2 1 1 d 1 s0 1 0 d 1 s4 1 5 dg y 5 s5 1 0 1 41d1 2 5 Ï46 i i 5 su u 2 1 u u 2 1 u u 2d1y2 v v1 v2 v3 5 fs12 1 12d 1 s22 1 12d 1 s02 1 02dg1y2 5 s2 1 5 1 0d1y2 5 Ï7 The distance between u and v is given by d su, vd 5 iu 2 vi 5 is1, 22 2 i, 4 2 5idi 5 fs12 1 02d 1 ss22d2 1 s21d2d 1 s42 1 52dg1y2 5 s1 1 5 1 41d1y2 5 Ï47.
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