Matrix exponential

Particular case of exponential map.
Given a matrix $K$ , we define a new matrix $\exp (K)$ as

\exp (K) = id + K + \frac{K^{2}}{2!} + \dots

It can be shown that it is well defined, i.e., this series is convergent. The proof involves the definition of a convenient norm in the matrix space, etcetera.

Matrix exponential has a bunch of good properties:

$\exp (0_{n}) = I_{n}$
$\exp (A + B) = \exp (A) \exp (B)$ provided $A B = B A$ . In the noncommutative case we have the Baker-Campbell-Hausdorff formula.
$\exp (A)^{- 1} = \exp (- A)$ for every $A \in M (n, R)$
$\exp (A^{T}) = (\exp (A))^{T}$ for every $A \in M (n, R)$
$\exp (P A P^{- 1}) = P \exp (A) P^{- 1}$ for every $A \in M (n, R)$ and $P \in G L (n, R)$
$\exp (A^{†}) = (\exp (A))^{†}$ for every $A \in M (n, R)$
If $A$ is a diagonal matrix with elements $λ_{i}$ , $e x p (A)$ is also diagonal, with elements $e^{λ_{i}}$ .
If

A = (\begin{array}{ccccc} 0 & 1 & 0 & \dots & 0 \\ 0 & 0 & 1 & \dots & 0 \\ ⋮ & ⋮ & ⋮ & ⋱ & ⋮ \\ 0 & 0 & 0 & \dots & 1 \\ 0 & 0 & 0 & \dots & 0 \end{array})

then

e^{A t} = (\begin{array}{ccccc} 1 & t & t^{2} / 2 & \dots & t^{n - 1} / (n - 1)! \\ 1 & t & \dots & ⋮ \\ 1 & \dots & t^{2} / 2 \\ ⋱ & t \\ ⋱ & 1 \end{array})

Proof
First, check that $A$ can be seen as the matrix of the derivative operator on polynomial-of-degree-less-than-n vector space, if we take the basis ${1, \frac{x}{1!}, \frac{x^{2}}{2!}, \frac{x^{3}}{3!}}$ . And the other matrix above is the matrix of the linear operator $H_{t} : p (x) \mapsto p (x + t)$ acting on the same vector space with the same basis.
Now, classical Taylor's theorem states that

e^{A t} (p (x)) = (id + t A + t^{2} \frac{A^{2}}{2!} + \dots) (p (x)) = p (x) + t p^{'} (x) + t^{2} \frac{p^{″} (x)}{2!} + \dots = p (x + t) = H_{t} (p (x))

That is to say

e x p (A t) = H_{t}

$◼$

Given any $K \in M (n, R)$ , let $γ (t) = \exp (t K)$ for $t \in R$ . Observe that $γ$ is differentiable, and $γ^{'} (t)$ can be obtained by term by term differentiation in the infinite sum (it results from convergence theorems). In fact, $γ^{'} (t) = K γ (t)$ .
Moreover, we can check that $γ : R \mapsto G L (n, R)$ is a one-parameter subgroup with $γ^{'} (0) = K$
The surprise is that the converse is also true:

Proposition
If

γ : R \to G L (n, R)

is a one-parameter subgroup then

γ (t) = \exp (t A)

where $A = γ^{'} (0)$
$◼$

Proof
It is based on the existence and uniqueness theorem of ODEs. Let $y = γ (t) x$ , where $x$ is a constant vector in $R^{n}$ . Then, with the definition of the derivative we check that $y$ is a solution of

y^{'} = A y

with $A = γ^{'} (0)$ . But $e^{A t}$ is another one...
$◼$
Observe that $γ (R)$ could be contained in any subgroup $G$ of $G L (n, R)$ . In this case, $A \in g$ . On the contrary, if we take any $A \in g$ , then $\exp (t A) \in G$ (proof?).

Thus, if we have a matrix $M_{τ}$ depending on the parameter $τ$ in such a way that ${M_{τ}}$ constitutes a one-parameter group then

M_{τ} = id + K τ + K^{2} \frac{τ^{2}}{2!} + \dots = \exp (τ K); τ \in R

for some matrix $K$ not depending on $τ$ .

Another important result, although more general that matrices groups, is:

Proposition
Given $F : G_{1} ⟼ G_{2}$ a differentiable group homomorphism, then

F (\exp (A)) = \exp (d F (A))

$◼$