Lecture 11

[[lecture-data]]

2024-09-20

Readings

2. Chapter 2

Claim: AMn is unitary if and only if there exists a BMn skew hermitian such that A=eB

Skew hermitian means that B=B. In particular, B is normal. Lets say that we unitarily diagonalize B so we can write B=UDU where UMn is unitary and DMn is diagonal. Then

B=B(UDU)=UDUUDU=UDUD=D

And this implies that the elements of D are purely imaginary.

Matrix exponential

The matrix exponential is

eB=i=01i!Bi
Theorem

AMn is unitary if and only if there exists a BMn skew hermitian such that A=eB
( see a matrix is unitary if and only if it equals the exponential of a skew hermitian matrix)

Proof

Suppose B is skew hermitian and A=eB. B is normal and we can say B=UDU where U is unitary and D is diagonal. Recall that the elements of the diagonal of D are pure imaginary. ie, there exist θ1,,θnR such that >$$\begin{bmatrix} \theta_{1} & 0 & \dots & 0 \
0 & \theta_{2} & \ddots & \vdots & \
\vdots & \ddots & \ddots & 0 \
0 & \dots & 0 & \theta_{n}
\end{bmatrix}$$
Then we have

A=eB=eUDU=UeDU=U[eθ1000eθ2000eθn]U

where and each of the matrices in the last product are unitary! And so A is unitary, since the product of unitary matrices is unitary

Now, suppose A is unitary. So A is normal, say A=WΩW. Then

Ω=[ω1000ω2000ωn]

And since A is unitary, it is an isometry, which implies that each of the eigenvalues of A have modulus 1 (||ωi||=1i). Thus, we can express each ωi as an angle between it and the real axis and this is given by eδi for δiR. So

Ω=[eδi000eδi000eδi]

Define

Σ=[δ1000δ2000δn]

Then B=WΣW is skew hermitian ! Then

eB=eWΣW=WeΣW=WΩW=A

3. Chapter 3 : Jordan Decomposition

Classifying matrices according to equivalence classes of similarity.

Jordan Block

A is a k×k Jordan block (denoted Jk(λ)) is

[λ1000λ1λ10λ100λ]

And this has eigenvalue λ with algebraic multiplicity k. The eigenvectors
x:[λIJk(λ)]x=0 can be found to be x1=anything and x2==xn=0, so the eigenvector has geometric multiplicity 1.

(see Jordan block)

When the eigenvalue of zero, when we raise it to the order of the block k, we get back the 0th matrix. We can see that

[Jk(0)]ij={1if =ji0otherwise

and rank[Jk(0)]=(k)0

Jordan Matrix

A Jordan Matrix is the direct sum of Jordan blocks.

J=Jn1(λ1)Jn2(λ2)Jns(λs)=[Jn1(λ1)000Jn2(λ2)000Jns(λn)]

"=i=1sJni(λi)" and λi not necessarily distinct.

This is a block diagonal matrix with the jordan blocks along the diagonal.

(see Jordan Matrix)

Theorem

For each AMn there exists a Jordan Matrix J such that A is similar to J:

A=SJS1

for SMn. Further, J is unique up to reordering its Jordan blocks.

Example

AM3 with all eigenvalues π has 3 possible similarity classes, one for each size of largest Jordan blocks.

(see Jordan's Theorem)

Note

A matrix is diagonalizable if and only if each of its jordan blocks are of size 1.