18.06.28: Complex vector spaces Lecturer: Barwick I worked so hard to understand it that it must be true. — James Richardson 18.06.28: Complex vector spaces From last time … I was alluding to a way to make complex multiplication easier to understand. The idea is this: for any 푧 = 푎 + 푏푖 ∈ C, you may consider the matrix 푎 −푏 푀 = ( ). 푧 푏 푎 On the newest problem set, you’ll show that addition of complex numbers is addition of these matrices, multiplication of complex numbers is multiplica- tion of these matrices (!), and one more thing … 18.06.28: Complex vector spaces Complex conjugation is the map 푧 푧 that carries 푧 = 푎 + 푏푖 to 푧 = 푎 − 푏푖. ⊺ You’ll see that 푀푧 = 푀푧. Complex conjugation can be used to extract the real and complex parts of your complex number: 2푎 = 푧 + 푧; 2푏푖 = 푧 − 푧. 18.06.28: Complex vector spaces Complex conjugation also gives you the length of the vector ~푣 ∈ R2 corre- sponding to 푧 ∈ C: ‖~푣‖2 = 푧푧. So 푧 = √푧푧 exp(푖휃) for some (and hence infinitely many) 휃 ∈ R. If 푧 ≠ 0, there is a unique such 휃 ∈ [0, 2휋); this is sometimes called the argument of 푧, but it’s annoying to write down a good formula. It’s better to think of it as an element of R/2휋Z. 18.06.28: Complex vector spaces One last general thing about the complex numbers, just because it’s so impor- tant. Theorem (“Fundamental theorem of algebra”). For any polynomial 푛 푖 푓(푧) = ∑ 훼푖푧 푖=0 with complex coefficients 훼푖 such that 훼푛 ≠ 0, there exist complex numbers 푤1,…, 푤푛 such that 푓(푧) = 훼푛(푧 − 푤1)(푧 − 푤2) ⋯ (푧 − 푤푛). This is actually a theorem of topology, not algebra, but there you go. 18.06.28: Complex vector spaces Here’s a cool example: 푓(푧) = 푧푛 − 1. Let’s find the roots! 18.06.28: Complex vector spaces The set C푛 is the set of column vectors 푧1 푧 푣 = ( 2 ) ⋮ 푧푛 with 푧푖 ∈ C. One can add such vectors componentwise, and one can multiply any such vector with a complex scalar. So this is the fundamental example of a complex vector space. 18.06.28: Complex vector spaces Note that R푛 ⊂ C푛. Now if 푣 ∈ C푛, then 푣 = 푣 if and only if 푣 ∈ R푛. 18.06.28: Complex vector spaces More generally, a complex vector subspace 푉 ⊆ C푛 is a subset such that: (1) for any 푣, 푤 ∈ 푉, one has 푣 + 푤 ∈ 푉; (2) for any 푣 ∈ 푉 and any 푧 ∈ C, one has 푧푣 ∈ 푉. 18.06.28: Complex vector spaces 푛 Vectors 푣1,…, 푣푘 span a vector subspace 푉 ⊆ C over C if and only if every vector 푤 ∈ 푉 can be written as a C-linear combination of the 푣푖, i.e., 푘 푤 = ∑ 푧푖푣푖, 푖=1 where each 푧푖 ∈ C. 18.06.28: Complex vector spaces Similarly, the vectors 푣1,…, 푣푘 are linearly independent over C if and only if any vanishing C-linear combination 푘 ∑ 푧푖푣푖 = 0 푖=1 is a trivial C-linear combination, so that 푧1 = ⋯ = 푧푘 = 0. A C-basis of 푉 is thus a collection of vectors of 푉 that is linearly independent over C and spans 푉 over C. 18.06.28: Complex vector spaces Let’s do an example to appreciate the distinction. Let’s think of C2, and let’s think of the complex line 푧 퐿 = {( ) ∈ C2 | 3푧 − 2푤 = 0} ⊂ C2. 푤 Now C2 ≅ R4, so that complex line is a real plane: { 푧1 } { | } { 푧2 | 3푧1 − 2푤1 = 0 } 퐿 = ( ) ∈ R4 | ⊂ R4. { | } { 푤1 3푧2 − 2푤2 = 0 } { 푤2 | } 18.06.28: Complex vector spaces 2 The single vector ( ) forms a C-basis of 퐿. 3 2푖 Another legit C-basis would be the single vector ( ). 3푖 The vectors 2 0 { } { 0 2 } ( ),( ) { } { 3 0 } { 0 3 } forms an R-basis of 퐿 over R. 18.06.28: Complex vector spaces Another example: consider the complex vector subspace 푊 ⊂ C2 spanned by 푖 ( ). 1 Here’s an important sentence to parse correctly: 푊 does not have a C-basis consisting of real vectors. 0 −1 1 0 A real basis for 푊 ⊂ R4 consists of ( ) and ( ) 1 0 0 1 18.06.28: Complex vector spaces Proposition. Any complex vector subspace 푊 ⊂ C푛 of complex dimension 푘 has an underlying real vector space of dimension 2푘. To see why, take a C-basis {푤1,…, 푤푘} of 푊. Now {푤1, 푖푤1,…, 푤푘, 푖푤푘} is an R-basis of 푊. 18.06.28: Complex vector spaces In the other direction, a real vector subspace 푉 ⊆ R푛 generates a complex 푛 vector subspace 푉C ⊆ C , called the complexification; this is the set of all C-linear combinations of elements of 푉: 푘 푛 푉C ≔ {푤 ∈ C | 푤 = ∑ 훼푖푣푖, for some 훼1,…, 훼푘 ∈ C, 푣1,…, 푣푘 ∈ 푉}. 푖=1 Note that not all complex vector subspaces of C푛 are themselves complexifi- 푖 cations; the complex vector subspace 푊 ⊂ C2 spanned by ( ) provides a 1 counterexample. (A complex vector space is a complexification if and only if it has a C-basis consisting of real vectors.) 18.06.28: Complex vector spaces Now, most importantly, we may speak of complex matrices (i.e., matrices with complex entries). All the algebra we’ve done with matrices over R works perfectly for matrices over C, without change. 18.06.28: Complex vector spaces However, the freedom to contemplate complex matrices offers us new hori- zons when it comes to questions about eigenspaces and diagonalization. Let’s contemplate the matrix 0 −1 퐴 = ( ). 1 0 2 The characteristic polynomial 푝퐴(푡) = 푡 + 1 doesn’t have any real roots, so there’s no hope of diagonalizing 퐴 over R. Over C, however, we find eigenvalues 푖, −푖. Let’s try to diagonalize 퐴. 18.06.28: Complex vector spaces 푖 1 Let’s begin with 퐿 = ker(푖퐼 − 퐴) = ker ( ). It’s dimension 1, and it’s 푖 −1 푖 1 spanned by the vector ( ). −푖 −푖 1 1 And 퐿 = ker ( ) is dimension 1 and spanned by ( ). −푖 −1 −푖 푖 18.06.28: Complex vector spaces Note that neither 퐿푖 nor 퐿−푖 is a complexification. However, we do have a basis 1 1 {( ),( )} of C2 consisting of eigenvectors of 퐴, and writing 푇 in −푖 푖 퐴 terms of this basis gives us the matrix 푖 0 ( ). 0 −푖 So 퐴 is not diagonalizable over R, but it is diagonalizable over C. 18.06.28: Complex vector spaces There’s one more new thing you can do with a complex matrices that doesn’t quite work for real matrices: you can conjugate their entries. Of particular import is the conjugate transpose: ⊺ 퐴∗ ≔ (퐴) = (퐴⊺). We’ll understand the significance of this operation next time..