Welcome to mathimatikoi.org forum; Enjoy your visit here.

Matrix exponential

Linear Algebra
Post Reply
User avatar
Grigorios Kostakos
Founder
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Matrix exponential

#1

Post by Grigorios Kostakos » Tue Dec 15, 2015 7:33 am

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
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Matrix exponential

#2

Post by Papapetros Vaggelis » Tue Dec 15, 2015 7:35 am

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

\(\displaystyle{b_{a\,a}=1\,,b_{a\,j}=0\,,j\in\left\{1,...,n\right\}-\left\{a\right\}\,,a\in\left\{1,2,...,n\right\}}\).
User avatar
Grigorios Kostakos
Founder
Founder
Articles: 0
Posts: 460
Joined: Mon Nov 09, 2015 1:36 am
Location: Ioannina, Greece

Re: Matrix exponential

#3

Post by Grigorios Kostakos » Tue Dec 15, 2015 7:39 am

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
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Matrix exponential

#4

Post by Papapetros Vaggelis » Fri Dec 18, 2015 5:53 pm

Some comments :

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

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

Here is a useful link
Papapetros Vaggelis
Community Team
Community Team
Articles: 0
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Matrix exponential

#5

Post by Papapetros Vaggelis » Mon Dec 21, 2015 4:35 pm

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 :

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