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## Matrix exponential

Linear Algebra
Grigorios Kostakos
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### Matrix exponential

Let $A$ a $n\times{n}$-matrix over $\mathbb{K}$ ( $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ ). We define $e^{A}:=\displaystyle\mathop{\sum}\limits_{k=0}^{+\infty}{\frac{A^k}{k!}}\,.\quad{(\dagger)}$ (1) If $A,P\in{\cal{M}}_{n\times{n}}(\mathbb{K})$ and $P$ is invertible, prove that $e^{PAP^{-1}}=P\,e^A\,P^{-1}\,.$ (2) If $\Delta=(\delta_{ij})\,,\quad\delta_{ij}= \begin{cases} 0\,,&i\neq j\\ \delta_{ii}\,, &i=j \end{cases}$ is a diagonal $n\times{n}$-matrix over $\mathbb{K}$, prove that $e^{\Delta}=\left({\begin{array}{cccc} e^{\delta_{11}} & 0 & \cdots & 0\\ 0 & e^{\delta_{22}} & \cdots & 0\\ \vdots & \vdots & \ddots & \vdots\\ 0 & 0 & \cdots & e^{\delta_{nn}} \end{array}}\right)\,.$ $(\dagger)$ NOTE: The series converges for every $A\in{\cal{M}}_{n\times{n}}(\mathbb{K})$.
Grigorios Kostakos
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### Re: Matrix exponential

1) We will prove that $\displaystyle{\left(P\,A\,P^{-1}\right)^{k}=P\,A^{k}\,P^{-1}\,,k\in\mathbb{N}\cup\left\{0\right\}}$ .

For $\displaystyle{k=0}$ or $\displaystyle{k=1}$ , it's obvious.

For $\displaystyle{k=2}$

\displaystyle{\begin{aligned} \left(P\,A\,P^{-1}\right)^2&=\left(P\,A\,P^{-1}\right)\left(P\,A\,P^{-1}\right)\\&=\left[\left(P\,A\,P^{-1}\right)\,P\right]\left(A\,P^{-1}\right)\\&=\left[\left(P\,A\right)\left(P^{-1}\,P\right)\right]\left(A\,P^{-1}\right)\\&=\left(P\,A\right)\left(A\,P^{-1}\right)\\&=P\,A^2\,P^{-1}\end{aligned}}

Let $\displaystyle{\left(P\,A\,P^{-1}\right)^{k}=P\,A^{k}\,P^{-1}\,,k\geq 2\,\,(I)}$

For $\displaystyle{k=k+1}$

\displaystyle{\begin{aligned} \left(P\,A\,P^{-1}\right)^{k+1}&=\left(P\,A\,P^{-1}\right)^{k}\,\left(P\,A\,P^{-1}\right)\\&\stackrel{(I)}{=}\left(P\,A^{k}\,P^{-1}\right)\,\left(P\,A\,P^{-1}\right)\\&=\left[\left(P\,A^{k}\,P^{-1}\right)\,P\right]\,\left(A\,P^{-1}\right)\\&=\left[\left(P\,A^{k}\right)\,\left(P^{-1}\,P\right)\right]\,\left(A\,P^{-1}\right)\\&=\left(P\,A^{k}\right)\left(A\,P^{-1}\right)\\&=P\,A^{k+1}\,P^{-1}\end{aligned}}

So,

\displaystyle{\begin{aligned} e^{P\,A,P^{-1}}&=\sum_{k=0}^{+\infty}\frac{\left(P\,A\,P^{-1}\right)^{k}}{k!}\\&=\sum_{k=0}^{+\infty}\frac{P\,A^{k}\,P^{-1}}{k!}\\&=P\left(\sum_{k=0}^{+\infty}\frac{A^{k}\,P^{-1}}{k!}\right)\\&=P\left(\sum_{k=0}^{+\infty}\frac{A^{k}}{k!}\right)\,P^{-1}\\&=P\,e^{A}\,P^{-1}\end{aligned}}

2) $\displaystyle{\Delta^{k}=\left(d_{i\,j}\right)\,\,\,\,,d_{i\,j}=\begin{cases} 0\,\,\,\,,i\neq j\\ \left(\delta_{i\,i}\right)^{k}\,\,\,\,\,\,\,,i=j \end{cases}}$

Therefore,

\displaystyle{\begin{aligned} e^{\Delta}&=\sum_{k=0}^{+\infty}\frac{\Delta^{k}}{k!}\\&=\lim_{k\to +\infty}\sum_{m=0}^{k}\frac{\Delta^{m}}{m!}\\&=\lim_{k\to +\infty}\sum_{m=0}^{k}\left(\frac{\delta_{1\,1}^{m}}{m!}\,B_{1}+...+\frac{\delta_{n\,n}^{m}}{m!}\,B_{n}\right)\\&=\left(\lim_{k\to +\infty}\sum_{m=0}^{k}\frac{\delta_{1\,1}^{m}}{m!}\right)\,B_{1}+...+\left(\lim_{k\to +\infty}\sum_{m=0}^{k}\frac{\delta_{n\,n}^{m}}{m!}\right)\,B_{n}\\&=e^{\delta_{1\,1}}\,B_{1}+...+e^{\delta_{n\,n}}\,B_{n}\end{aligned}}

where,

$\displaystyle{B_{a}=\left(b_{i\,j}\right)}$ is a diagonial $\displaystyle{n\times n}$ matrix and

$\displaystyle{b_{a\,a}=1\,,b_{a\,j}=0\,,j\in\left\{1,...,n\right\}-\left\{a\right\}\,,a\in\left\{1,2,...,n\right\}}$.
Grigorios Kostakos
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### Re: Matrix exponential

A short cut for $\bigl({PAP^{-1}}\bigr)^k=PA^{k}P^{-1}\,, \; k\in\mathbb{N}\cup\{0\}\,,$ is
$\bigl({PAP^{-1}}\bigr)^k=\mathop{\underbrace{PAP^{-1}PAP^{-1}\cdots PAP^{-1}}}\limits_{k-{times}}=PAI_{n}AI_{n}\cdots I_{n}AP^{-1}=PA^{k}P^{-1}\,,$ where $I_n$ is the $n\times n$ identity matrix.
Grigorios Kostakos
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### Re: Matrix exponential

The convergence of the series $\displaystyle{\sum_{n=0}^{\infty}\dfrac{A^{n}}{n!}}$, where $\displaystyle{A\in\mathbb{M}_{n}(\mathbb{R})}$ ,

is taken with respect to the $\displaystyle{\rm{Frobenious}}$ norm :

$\displaystyle{||\cdot||:\mathbb{M}_{n}(\mathbb{R})\longrightarrow \mathbb{R}\,,||A||=\sqrt{\rm{Tr}(A\,A^{t})}}$ .

Then, $\displaystyle{\left(\mathbb{M}_{n}(\mathbb{R}),+,\cdot\right)\simeq \left(\mathbb{B}(\mathbb{R}^n,\mathbb{R}^n),+,\cdot\right)}$

as real $\displaystyle{\rm{Banach}}$ - spaces. We define

$\displaystyle{T:\mathbb{M}_{n}(\mathbb{R})\longrightarrow \mathbb{B}(\mathbb{R}^n,\mathbb{R}^n)\,,T(A)(x)=A\,x^{t}\,,\forall\,A\in\mathbb{M}_{n}(\mathbb{R})\,,\forall\,x\in\mathbb{R}^n}$ .

Papapetros Vaggelis
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### Re: Matrix exponential

Hello Grigorios.

Here is another proof of the equation

$\displaystyle{e^{P\,A\,P^{-1}}=P\,e^{A}\,P^{-1}}$ , where $\displaystyle{P\in GL_{n}(\mathbb{R})}$ .

Proof

We define $\displaystyle{f:\mathbb{M}_{n}(\mathbb{R})\longrightarrow \mathbb{M}_{n}(\mathbb{R})}$

by $\displaystyle{f(A)=P\,A\,P^{-1}}$.

For every $\displaystyle{A\,,B\in\mathbb{M}_{n}(\mathbb{R})}$ holds :

$\displaystyle{f(A+B)=P\,(A+B)\,P^{-1}=P\,A\,P^{-1}+P\,B\,P^{-1}=f(A)+f(B)}$

$\displaystyle{f(A\,B)=P\,(A\,B)\,P^{-1}=(P\,A\,P^{-1}(P\,B\,P^{-1})=f(A)\,f(B)}$

$\displaystyle{f(I_{n})=I_{n}}$

$\displaystyle{f(c\,A)=P\,(c\,A)\,P^{-1}=c\,P\,A\,P^{-1}=c\,f(A)\,,\forall\,c\in\mathbb{R}}$ .

So, the function $\displaystyle{f}$ is $\displaystyle{\mathbb{R}}$ - linear (and ring homomorphism) .

Also, for every $\displaystyle{A\in\mathbb{M}_{n}(\mathbb{R})}$ holds :

$\displaystyle{||f(A)||=||P\,A\,P^{-1}||\leq ||P||\,||A||\,||P^{-1}||=||P||\,||P^{-1}||\,||A||}$

which means that $\displaystyle{f}$ is bounded, that is, it is continuous function with respect to

the metric generated by $\displaystyle{\rm{Frobenious}}$ norm.

Let $\displaystyle{A\in\mathbb{M}_{n}(\mathbb{R})}$ . In the above metric space :

$\displaystyle{\left(B_{n}=\sum_{k=0}^{n}\dfrac{A^{k}}{k!}\right)_{n\in\mathbb{N}}\longrightarrow e^{A}}$

so :

$\displaystyle{\left(f(B_{n})\right)_{n\in\mathbb{N}}\longrightarrow f(e^{A})=P\,e^{A}\,P^{-1}}$

where :

$\displaystyle{f(B_{n})=P\,B_{n}\,P^{-1}=P\,\sum_{k=0}^{n}\dfrac{A^{k}}{k!}\,P^{-1}=\sum_{k=0}^{n}\dfrac{(P\,A\,P^{-1})^{k}}{k!}}$

On the other hand,

\displaystyle{\begin{aligned} e^{P\,A\,P^{-1}}&=\lim_{n\to +\infty}\sum_{k=0}^{n}\dfrac{(P\,A\,P^{-1})^{k}}{k!}\\&=\lim_{n\to +\infty}f(B_{n})\end{aligned}}

Since the limit is unique, we get :

$\displaystyle{e^{P\,A\,P^{-1}}=f(e^{A})=P\,e^{A}\,P^{-1}}$ .