The matrix exponential is the exponential [function of an operator, $X,$](Functions%20of%20linear%20operators.md) such that $f(X)=e^{X}.$
We generalize slightly by including a parameter $t$ such that $t\in\mathbb{R}$ so that $f(X)=e^{tX}$
In addition we also define a [complex matrix exponential.](Complex%20matrix%20exponentials.md) 
# Power series representation of matrix exponentials
Given an operator $X$ or a matrix $X$ with real or complex elements, in order to [evaluate](Matrix%20exponentials.md#Evaluating%20matrix%20exponentials) a matrix exponential, we need to first write it as a [power series](Functions%20of%20linear%20operators.md#as%20a%20power%20series),
$e^{tX} = \sum_{n=0}^{\infty} \frac{(tX)^n}{n!}.$
which follows from the expression of any function, $e^x$ [as a power series](Analysis%20(index).md#As%20a%20power%20series).
This expression provides a general for step for [evaluating matrix exponentials.](Matrix%20exponentials.md#Evaluating%20matrix%20exponentials)
# Matrix exponentials as limits of sequences
We may expand matrix exponentials such that
$e^{tX} = \lim_{t\rightarrow 0}\big(1+tX\big)^{1/t}.$
This follows from the representation of $e^x$ [as a limit of a sequence](Analysis%20(index).md#As%20a%20limit%20of%20a%20sequence).
An example where this is used is in the construction of some solutions to differential equations.
%%which solutions to differential equations is this used for? Please add more notes about diff eqs where this may be considered.%%
# Properties of matrix exponentials
1) $e^0 = \mathbb{1}$ ^84e328
2) $(e^{X})^\dagger = e^{X^\dagger}$ ^799851
3) $e^X$ is invertible and $(e^{X})^{-1} = e^{-X}$ ^d3c55b
4) $e^{(\alpha + \beta)X} = e^{\alpha X}e^{\beta X}$, $\forall \alpha, \beta \in \mathbb{C}$ ^97521c
5) $[X,Y] = 0$ implies that $e^{X+Y} = e^Xe^Y = e^Ye^X$ ^4f36a0
6) Given an invertible matrix $C$, $e^{CXC^{-1}}=Ce^XC^{-1}$ ^819683
7) $\frac{d}{dt}e^{At} = Ae^{At} = e^{At}A$ ^2b1be7
8) $\frac{d}{dt}e^{At}e^{Bt} = Ae^{At}e^{Bt} + e^{At}Be^{Bt} = e^{At}(A+B)e^{Bt}$. Where we should be careful to note that that $e^{At}(A+B)e^{Bt} \neq (A+B)e^{At}e^{Bt} \neq e^{At}e^{Bt}(A+B)$ ^d965bd
Property 1. can be shown by direct substitution, 2. follows from the definition of the [adjoint,](Adjoint.md) and 3. from the fact that $e^{-X}e^{X}=e^0=1.$ Property 4. is a special case of 5. Thus, properties 4. and 5. are [provable](Matrix%20exponentials.md#Proof%20of%20properties%204%20Matrix%2020exponentials%20md%2097521c%20and%205%20Matrix%2020exponentials%20md%204f36a0%20of%20matrix%20exponentials) by taking the [Taylor series](Matrix%20exponentials.md#Power%20series%20representation%20of%20matrix%20exponentials) of $e^{X+Y}$ and evaluating it. 6. is [proven](Matrix%20exponentials.md#Proof%20of%20property%206%20Matrix%2020exponentials%20md%20819683%20of%20matrix%20exponentials) using the property $(CXC^{-1})^m = CX^mC^{-1}$. Properties 7. and 8. can be [proven](Matrix%20exponentials.md#Proof%20of%20properties%207%20Matrix%2020exponentials%20md%202b1be7%20and%208%20Matrix%2020exponentials%20md%20d965bd%20of%20matrix%20exponentials) by applying the elementary definition of the [derivative.](Analysis%20(index).md#Derivatives)
%%Don't just say 5 is provable evaluate it below?%%
%%What is the name of the property $(CXC^{-1})^m = CX^mC^{-1}$?%%
### Derivatives of matrix exponentials
$e^{tX}$ is a [smooth function](Smooth%20function.md) in a [complex vector space](Complex%20vector%20spaces.md) (Along with the [complex matrix exponentials](Complex%20matrix%20exponentials.md)), $\mbox{M}_n(\mathbb{C})$, meaning that it's differentiable such that
$\frac{d}{dt}e^{tX} = Xe^{tX}$ where in particular at $t=0$, the derivative yields an element of a [Lie algebra](Lie%20algebras.md) for a corresponding matrix exponential that is an element of a [Lie group.](Lie%20groups.md)
$\frac{d}{dt}e^{tX}|_{t=0} = X$
# Role of matrix exponentials in Lie groups and Lie algebras
The matrix exponential is the _mechanism for passing information from a [[Lie algebras]] to its Lie Group_. That is if $X\in\mathfrak{g}$ we have [Lie groups](Lie%20groups.md) elements $e^{tX}$, $\forall t \in \mathbb{R}$. We use the form of a [[Complex matrix exponentials]], $e^{itX}$ for Lie Groups, where $iX\in\mathfrak{g} \,\,\, \forall X\in\mathfrak{g}$. ^362102
This role in Lie Groups and Algebras depends on the properties of the [derivatives of matrix exponentials.](Matrix%20exponentials.md#Derivatives%20of%20matrix%20exponentials)
## Algebraic properties matrix exponentials
It is _not_ always true that $e^{X+Y}=e^Xe^Y$ and this fact is behind several emergent identities that are proven by considering the [role of matrix exponentials in Lie groups and Lie algebras](Matrix%20exponentials.md#Role%20of%20matrix%20exponentials%20in%20Lie%20groups%20and%20Lie%20algebras):
* The [Lie product formula](Lie%20product%20formula.md)
* A very similar related identity is the [Trotter Product Formula](Lie%20product%20formula.md#Trotter%20Product%20Formula)
* The [Baker-Campbell-Hausdorff lemma](Baker-Campbell-Hausdorff%20lemma.md)
* The [Baker-Campbell-Hausdorff formula](Baker-Campbell-Hausdorff%20formula.md)
# [Diagonalizing](Diagonalizable%20matrices.md) matrix exponentials
For any $X\in\mbox{M}_n(\mathbb{C})$, if $X$ is diagonalizable with [eigenvalues](Eigenvalues%20and%20eigenvectors.md) $\lambda_1 ... \lambda_n$ then $e^X$ is also diagonalizable with eigenvalues $e^{\lambda_1}...e^{\lambda_n}$.
Since the matrix exponential is diagonalizable we also conclude that $\det{e^X} = e^{\mbox{tr}{X}}.$ This follows directly from [properties 1 and 2](Diagonalizable%20matrices.md#Properties%20of%20diagonalizable%20matrices) of diagonalizable matrices where we find that
$\mathrm{det}({e^X})=\prod_i e^{\lambda_i}=e^{\sum_i\lambda_i}=e^{\mbox{tr}{X}}$ ^af041f
# Evaluating matrix exponentials
The most general way of evaluating a [matrix exponential](Matrix%20exponentials.md) is by approximating it through its [power series representation](Matrix%20exponentials.md#Power%20series%20representation%20of%20matrix%20exponentials). However, there are cases where it can be done exactly with fewer computations as shown below.
## Evaluating matrix exponentials using nilpotent matrices
Finding [nilpotent matrices](nilpotent%20matrices) is useful for truncating the exponential infinite series. In fact every matrix $X$ can be written as $X = S+N$ where $N^k = 0$ for some $k$.
Next notice that $[S,N] = 0$, meaning $e^X = e^{N+S} = e^{N}e^{S}$
## Evaluating matrix exponentials when $e^X$ is diagonalizable
Given $(CXC^{-1})^m = CX^mC^{-1}$ we can find $C$ in order to evaluate the matrix exponential, allowing us to decompose the exponential into $e^X = C \begin{pmatrix} e^{\lambda_1} & & \\ & \ddots & \\ & & e^{\lambda_n}\end{pmatrix}C^{-1}$ in cases where the matrix exponential is [diagonalizable.](Matrix%20exponentials.md#Diagonalizing%20Diagonalizable%2020matrices%20md%20matrix%20exponentials)
---
# Proofs and examples
## Proof of convergence of the Power Series
Here we prove the [power series representation of matrix exponentials](Matrix%20exponentials.md#Power%20series%20representation%20of%20matrix%20exponentials) converges.
Define the norm as $||X||=(\sum_{j,k=1}^{n}|X_{jk}|^2)^{1/2} = (\mbox{tr}{X^{\dagger}X})^{1/2}$
From the [triangle inequality](Triangle%20inequality.md), $||X+Y|| = \sum_{j,k=1}^{n}|(X+Y)_{jk}|^{1/2} \leq ||X||+||Y||$
And from the [Cauchy-Schwarz inequality](Cauchy-Schwarz%20inequality.md) we know that $||X^m||\leq||X||^m$
Therefore, since $e^0 = \mathbb{1}$ we may write $e^{X} = \sum_{n=0}^{\infty} ||\frac{X^n}{n!}|| \leq ||I||+\sum_{n=1}^{\infty} \frac{||X||^n}{n!}<\infty$
meeting the criteria for _absolute convergence_.
## Proof of properties [4](Matrix%20exponentials.md#^97521c) and [5](Matrix%20exponentials.md#^4f36a0) of matrix exponentials
## Proof of property [6](Matrix%20exponentials.md#^819683) of matrix exponentials
## Proof of properties [7](Matrix%20exponentials.md#^2b1be7) and [8](Matrix%20exponentials.md#^d965bd) of matrix exponentials
## Example evaluations of matrix exponentials
### Example 1
Given $X_1=\begin{pmatrix} 0 & -a \\
a & b
\end{pmatrix},$
we find its characteristic polynomial to be $\lambda^2 + a^2 = 0$ yielding eigenvalues $\lambda = \pm ia$.
Next $C$ is constructed by finding the corresponding eigenvectors and taking them to be the columns of $C$:
$\begin{pmatrix} 0 & -a \\
a & b
\end{pmatrix}\begin{pmatrix} x \\
y
\end{pmatrix} = \pm ia \begin{pmatrix} x \\
y
\end{pmatrix}$
The respective eigenvectors are $\begin{pmatrix} i\\
1
\end{pmatrix}$
and $\begin{pmatrix} 1\\
i
\end{pmatrix}$
with which we find $C = \begin{pmatrix} i & 1 \\
1 & i
\end{pmatrix} \;\; \textrm{and} \;\;\; C^{-1} = \begin{pmatrix} 1/2 & -i/2 \\
-i/2 & 1/2
\end{pmatrix}$
where we recall the formula for the [operator inverse](Operator%20inverse.md). Finally,
$e^{X_1} = C \begin{pmatrix} e^{\lambda_1} & 0 \\
0 & e^{\lambda_2}
\end{pmatrix} C^{-1} = \begin{pmatrix} i & 1 \\
1 & i
\end{pmatrix}\begin{pmatrix}e^{-ia} & 0 \\
0 & e^{ia}
\end{pmatrix}\begin{pmatrix} 1/2 & -i/2 \\
-i/2 & 1/2
\end{pmatrix}$
$= \begin{pmatrix} (e^{-ia}+e^{ia})/2 & i(e^{ia}-e^{-ia})/2 \\
i(e^{-ia}-e^{ia})/2 & (e^{-ia}+e^{ia})/2
\end{pmatrix} = \begin{pmatrix} \cos(a) & -\sin(a) \\
\sin(a) & \cos(a)
\end{pmatrix}$
### Example 2
Given $X_2=\begin{pmatrix} 0 & a & b\\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix}$
We note that we quickly arrive at a nilpotent matrix:
$(X_2)^2=\begin{pmatrix} 0 & a & b\\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix} 0 & a & b\\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix} = \begin{pmatrix} 0 & 0 & ba+cb\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}$
$(X_2)^3 = \begin{pmatrix} 0 & 0 & ba+cb\\
0 & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}\begin{pmatrix} 0 & a & b\\
0 & 0 & c \\
0 & 0 & 0
\end{pmatrix} = 0$
Thus the Taylor expansion truncates at the 2nd order, and we can use this to evaluate exactly:
$e^{X_2} = I + X_2 + \frac{(X_2)^2}{2!} = \begin{pmatrix} 1 & a & (ba+cb)/2\\
0 & 1 & c \\
0 & 0 & 1
\end{pmatrix}$
### Example 3
Given
$X_3 = \begin{pmatrix} a & b \\
0 & a
\end{pmatrix}$
We can clearly see that, as with all $2\times2$ upper triangular matrices, we can quickly decompose it into a sum of a diagonal matrix and a nilpotent matrix:
$X_3 = \begin{pmatrix} a & 0 \\
0 & a
\end{pmatrix}+\begin{pmatrix} 0 & b \\
0 & 0
\end{pmatrix}$
Therefore:
$e^{X_3} = e^{\begin{pmatrix} a & 0 \\
0 & a
\end{pmatrix}}e^{\begin{pmatrix} 0 & b \\
0 & 0
\end{pmatrix}} = \begin{pmatrix} e^a & be^a \\
0 & e^a
\end{pmatrix}$
---
# Recommended reading
Considering its central role in [Lie groups](Lie%20groups.md) and [Lie algebras](Lie%20algebras.md), a comprehensive view of the properties of matrix exponentials may be found in the following source:
* [Hall, B., _Lie Groups Lie Algebras and Representations_, Springer, 2nd edition, 2015.](Hall,%20B.,%20Lie%20Groups%20Lie%20Algebras%20and%20Representations,%20Springer,%202nd%20edition,%202015..md) (pgs 31-46.) This page draws heavily from this chapter.
Much of the same key pieces of information presented in the above source are also found in a similar context here:
* [Hall, Brian. _Quantum Theory for Mathematicians_, Springer (2013).](%5BGraduate%20Texts%20in%20Mathematics%20267%5D%20Brian%20C.%20Hall%20(auth.)%20-%20Quantum%20Theory%20for%20Mathematicians%20(2013,%20Springer-Verlag%20New%20York)%20-%20libgen.lc.pdf) This text also presents matrix exponentials as they relate to Lie Groups and Lie Algebras. The broader context of this work is as an introduction to quantum physics aimed at mathematicians.
%%rewrite this page as a page on _matrix operators,_ with _matrix exponentials as a subsection_%%
#MathematicalFoundations/Algebra/AbstractAlgebra/LinearAlgebra/Operators/Matrices
#MathematicalFoundations/Algebra/AbstractAlgebra/GroupTheory/Lie/LieGroups