Analysis MOC

Matrix exponential

The matrix exponential exp :ℂ𝑛×𝑛 →ℂ𝑛×𝑛 uses the power series definition of the exponential function on matrices. Let 𝐀 be a 𝑛 ×𝑛 real/complex matrix. Then 𝑒𝐀 is given by vec

𝑒𝐀=∞∑𝑘=0𝐀𝑘𝑘!

This is convergent for all 𝐀 ∈ℂ𝑛×𝑛 under any norm.

Proof of convergence

Let ‖ ⋅‖ denote the Operator norm. Then

0≤‖𝐀𝑘𝑘!‖=‖𝐀𝑘‖𝑘!≤‖𝐀‖𝑘𝑘!

Since 𝑒‖𝐴‖ =∑∞𝑘=0‖𝐴‖𝑘𝑘! converges, the sequence converges absolutely and uniformly by the Weierstraß M-test. ℂ𝑛×𝑛 is finite-dimensional: All finite dimensional normed vector spaces are Banach and All norms on a finite dimensional space are equivalent. Thus the series converges, and does so regardless of norm.

Properties

For any 𝐀 ∈ℂ𝑛×𝑛, the following properties hold: vec

1. For any invertible 𝐓 ∈GL(𝑛), 𝑒𝐓𝐀𝐓−1 =𝐓𝑒𝐀𝐓−1.
2. 𝑒𝑡𝐀 uniquely solves ˙𝐗(𝑡) =𝐀𝐗(𝑡) with initial condition 𝐗(0) =𝐈.
3. 𝑒𝑡𝐀𝑒𝑠𝐁 =𝑒𝑡𝐀+𝑠𝐁 if 𝐀𝐁 =𝐁𝐀 for all 𝑡,𝑠 ∈ℂ.
4. (𝑒𝐀)† =𝑒(𝐀†)
5. det𝑒𝐀 =𝑒tr⁡𝐀
6. 𝑒−𝑖𝜙⃗𝜎⋅⃗𝐧/2 =cos⁡(𝜙/2)𝐈 −𝑖sin⁡(𝜙/2)⃗𝜎 ⋅⃗𝐧 for ⃗𝐧 ∈𝕊3 (see Pauli matrices)

Proof of properties 1–5

Let 𝐓 ∈GL(n). Then

𝑒𝐓𝐀𝐓−1=∞∑𝑘=0(𝐓𝐀𝐓−1)𝑘𝑘!=∞∑𝑘=0𝐓𝐀𝑘𝐓−𝟏𝑘!=𝐓𝑒𝐀𝐓−1

proving ^P1

Let

𝐗(𝑡)=𝑒𝑡𝐀=∞∑𝑘=0(𝑡𝐀)𝑘𝑘!

Then

˙𝐗(𝑡)=∞∑𝑘=1𝑡𝑘−1𝐀𝑘(𝑘−1)!=𝐀∞∑𝑘=0(𝑡𝐀)𝑘𝑘!=𝐀𝐗(𝑡)

and by the Existence and uniqueness theorem for IVPs this is unique, proving ^P2.

By Binomial expansion

𝑒𝑡𝐀𝑒𝑠𝐁=(∞∑𝑘=0(𝑡𝐀)𝑘𝑘!)(∞∑ℓ=0(𝑠𝐁)ℓℓ!)=∞∑𝑘=0𝑘∑ℓ=0𝑡ℓ𝑠𝑘−ℓ𝐀ℓ𝐁𝑘−ℓℓ!(𝑘−ℓ)!=∞∑𝑘=0𝑘∑ℓ=0𝑘!ℓ!(𝑘−ℓ)!𝑡ℓ𝑠𝑘−ℓ𝐀ℓ𝐁ℓ−𝑘𝑘!=∞∑𝑘=0𝑘∑ℓ=0(𝑘ℓ)𝑡ℓ𝑠𝑘−ℓ𝐀ℓ𝐁ℓ−𝑘𝑘!=∞∑𝑘=0(𝑡𝐀+𝑠𝐁)𝑘𝑘!=𝑒(𝑡𝐀+𝑠𝐁)

proving ^P3

By basic properties of the Conjugate transpose

(𝑒𝐀)†=(∞∑𝑘=0𝐀𝑘𝑘!)†=∞∑𝑘=0(𝐀𝑘)†𝑘!=∞∑𝑘=0(𝐀†)𝑘𝑘!=𝑒(𝐀†)

proving ^P4.

Let 𝐃 be the Jordan canonical form so 𝐀 =𝐒𝐃𝐒−1. Then

det𝑒𝐀=det𝑒𝐃=𝑛∏𝑘=1𝑒𝐷𝑘𝑘=𝑒tr⁡𝐃=𝑒tr⁡𝐀

proving ^P5

Generalisations

• Vector flow
• Exponential map of Lie theory.

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