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Matrix exponential

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
I quote a question from Yahoo! Answers

Consider the following Matrix:
A =
[ 1 -1 0
1 3 0
4 6 -1 ]

(a) Find a Jordan form of the matrix, as well as a basis that corresponds to that Jordan form.
(b) Compute the exponential matrix eA.
I have given a link to the topic there so the OP can see my response.
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
$(a)$ The eigenvalues of $A=\begin{bmatrix}{1}&{-1}&{\;\;0}\\{1}&{\;\;3}&{\;\;0}\\{4}&{\;\;6}&{-1}\end{bmatrix}$ are: $$\det (A-\lambda I)=(-1-\lambda)(\lambda -2)^2=0\Leftrightarrow \lambda=-1\mbox{ (simple) }\vee \;\lambda=2\mbox{ (double)}$$ We have $\dim\ker (A+I)=1$ (simple eigenvalue) and $\dim \ker (A-2I)=3-\mbox{rank} (A-2I)=3-2=1$. So the canonical form of Jordan is $$J= \begin{bmatrix} {-1}&{0}&{0}\\{0}&{2}&{1}\\{0}&{0}&{2}\end{bmatrix} $$ A basis for $\ker (A+I)$ is $\{v=(0,0,1)^T\}$. Now, we need two linearly independent vectors $e_1,e_2$ such that $(A-2I)e_1=0$ and $(A-2I)e_2=e_1$. We easily find $e_1=(-3,3,2)^T$ and $e_2=(8,-5,0)^T$. As a consequence, the transition matrix $P$ satisfying $P^{-1}AP=J$ is $$P=[v\;\;e_1\;\;e_2]=\begin{bmatrix}{0}&{-3}&{\;\;8}\\{0}&{\;\;3}&{-5}\\{1}&{\;\;2}&{\;\;0}\end{bmatrix}$$

$(b)\;\;e^{A}=Pe^{J}P^{-1}=P\;\begin{bmatrix}{e^{-1}}&{0}&{0}\\{0}&{e^{2}}&{e^{2}}\\{0}&{0}&{e^{2}} \end{bmatrix}\;P^{-1}=\ldots=\begin{bmatrix}{e}&{e^{-1}}&{1}\\{e}&{e^3}&{1}\\{e^4}&{e^6}&{e^{-1}} \end{bmatrix}$
 
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