18.06.29: Complex matrices

Lecturer: Barwick

If it can be used again, it is not wisdom but theory. — James Richardson 18.06.29: Complex matrices

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. 18.06.29: Complex matrices

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.) 18.06.29: Complex matrices

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. 18.06.29: Complex matrices

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. 18.06.29: Complex matrices

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 18.06.29: Complex matrices

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. 18.06.29: Complex matrices

More generally, if we’re looking at a real matrix of the form

푎 −푏 푀 = ( ), 푏 푎 then 푀 = 푀푧 for 푧 = 푎 + 푏푖, and on the problem set, you’ll show that

2 푝푀(푡) = 푡 − (푧 + 푧)푡 + 푧푧.

The roots of this polynomial are 푧 and 푧. 18.06.29: Complex matrices

So, very pleasantly, 푀 is diagonalizable over C, and it’s similar to the matrix

푧 0 푀 = ( ). 0 푧 18.06.29: Complex matrices

This game of going back and forth between 푧 and 푀푧 is helpful in other ways. For example, let’s take a 2 × 3 complex matrix

1 2푖 2 + 3푖 퐴 = ( ). 1 − 4푖 5푖 1 − 푖

We can replace each complex entry 푧 with the 2×2 matrix 푀푧 that corresponds to it, giving us a 4 × 6 real matrix 푀퐴 … 18.06.29: Complex matrices

1 0 0 −2 2 −3 0 1 2 0 3 2 푀 = ( ). 퐴 1 4 0 −5 1 1 −4 1 5 0 −1 1

3 2 How is that helpful? Well, if we think of 푇퐴∶ C C given by multiplication by 퐴, we should be able to regard that as a linear map R6 R4 given by a

4 × 6 matrix. 푀퐴 is precisely that matrix! 18.06.29: Complex matrices

In particular, think about the transpose of 푀퐴. What complex matrix does it correpond to? 18.06.29: Complex matrices

Our last midterm is Friday. (sniff!)

▶ I know you’re sad, but try to work through the hurt.

▶ Five questions, as usual.

▶ It covers everything up to this page of the lectures.

▶ I’m aiming for a mean of around 90 again. I missed last time, but I suspect that had more to do with the shittiness of that particular week than with your ability to do the math. 18.06.29: Complex matrices

Back to our transpose:

1 0 1 −4 0 1 4 1 ( 0 2 0 5 ) 푀⊺ = ( ) . 퐴 ( −2 0 −5 0 ) 2 3 1 −1 ( −3 2 1 1 ) and let’s convert it back to a complex matrix … 18.06.29: Complex matrices

1 1 + 4푖 퐴∗ = ( −2푖 −5푖 ). 2 − 3푖 1 + 푖

This is the conjugate transpose of 퐴, so that

⊺ 퐴∗ = (퐴⊺) = (퐴) .

⊺ This clearly works in general, and we therefore find that 푀퐴 = 푀퐴∗ . 18.06.29: Complex matrices

A real matrix 퐴 is said to be symmetric if 퐴 = 퐴⊺.

A complex matrix 퐵 is said to be Hermitian if 푀퐵 is symmetric – or, equiva- lently, if 퐵 = 퐵∗.

(Note that a Hermitian matrix with real entries must be symmetric.)

We need to think about this a bit more carefully. For that, let’s contemplate the correct version of the dot product in C푛, and develop some notation. 18.06.29: Complex matrices

For 푣, 푤 ∈ C푛, write ⟨푣|푤⟩ ≔ 푣∗푤.

This is a complex number, called the inner product of two complex vectors; it extends the usual dot product, but notices that the linearity in the first coordi- nate is twisted: ⟨훼푣|푤⟩ = 훼⟨푣|푤⟩.

With this, one can repeat the usual definition of orthogonality with no prob- lem.

Lemma. An 푛 × 푛 complex matrix 퐵 is Hermitian if and only if, for any 푣, 푤 ∈ C푛, ⟨퐴푣|푤⟩ = ⟨푣|퐴푤⟩. 18.06.29: Complex matrices

Theorem (Spectral theorem; last big result of the semester). Suppose 퐵 a Her- mitian matrix. Then

(1) The eigenvalues of 퐵 are real.

(2) There is an orthogonal basis of eigenvectors for 퐵; in particular, 퐵 is diago- nalizable over C (and even over R if 퐵 has real entries).