Doubt in the defn of exponential operator.

Question

definition

Let $A$ be an $n\times n$ matrix. Then for $t\in \mathbb R$, $$e^{At}=\sum_{k=0}^\infty \frac{A^kt^k}{k!}\tag{1}$$

But in this definition, What they are meaning by the term $A^kt^k$, If I give this matrix $$A=\begin{pmatrix}1&-1\\2&3\end{pmatrix},$$

what can we say about the $(1)$, means How it will be?

Answer 1

One way to handle this kind of problem is to find the Jordan Normal Form of $A$: $$ A=\begin{bmatrix} -1-i&-1+i\\ 2&2 \end{bmatrix} \begin{bmatrix} 2-i&0\\ 0&2+i \end{bmatrix} \begin{bmatrix} -1-i&-1+i\\ 2&2 \end{bmatrix}^{-1} $$ Taking the exponential of $A$ is now pretty simple: $$ \begin{align} \exp(At)&= \begin{bmatrix} -1-i&-1+i\\ 2&2 \end{bmatrix} \begin{bmatrix} e^{2t}e^{-it}&0\\ 0&e^{2t}e^{it} \end{bmatrix} \begin{bmatrix} -1-i&-1+i\\ 2&2 \end{bmatrix}^{-1}\\ &=\begin{bmatrix} -1-i&-1+i\\ 2&2 \end{bmatrix} \begin{bmatrix} e^{2t}e^{-it}&0\\ 0&e^{2t}e^{it} \end{bmatrix} \begin{bmatrix} 2i&1+i\\ -2i&1-i \end{bmatrix}\\ &=e^{2t}\begin{bmatrix} \cos(t)-\sin(t)&-\sin(t)\\ 2\sin(t)&\cos(t)+\sin(t) \end{bmatrix} \end{align} $$

Answer 2

Answering your question $(1)$ will be

$$ e^{At} = \left[ \begin {array}{cc} {{\rm e}^{2\,t}}(\cos \left( t \right)- \sin \left( t \right)) &-{{\rm e}^{2\,t}}\sin \left( t \right) \\ 2\,{{\rm e}^{2\,t}}\sin \left( t \right) & {{\rm e}^{2\,t}}(\cos \left( t \right) +\sin\left( t \right)) \end {array} \right] .$$

Answer 3

To do exponentiation of matrices, the easiest way is to diagonalize the exponent first,

$U^{\dagger} AtU = D = \mbox{diag}\{D_1, D_2,...,D_n\} $ with eigen values $D_1,D_2,...,D_n$

since

$ U^{\dagger}exp(At) U = exp(D) = \mbox{diag}\{e^{D_1}, e^{D_2},...,e^{D_n} \} $ Then $$exp(At) = U exp(D)U^{\dagger} $$

For you matrix, you will get,

$$ e^{At} = \left( \begin{array}{cc} e^{2 t} \cos (t)-e^{2 t} \sin (t) & -e^{2 t} \sin (t) \\ 2 e^{2 t} \sin (t) & e^{2 t} \cos (t)+e^{2 t} \sin (t) \\ \end{array} \right) .$$