Given a solution to find a matrix

Question

For $e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & e^{2t}+e^{-t}\end{bmatrix}$ for all t $\in$ $\mathbb{R}$. how to find A?

Answer 1

The matrix exponential is given by:

$$\tag 1 e^{At} = \sum_{k=0}^{n-1} \alpha_k A^k$$

where the $\alpha_i$'s are determined from the set of equations given by the eigenvalues of A, as:

$$\tag 2 e^{\lambda_i t} = \sum_{k=0}^{n-1} \alpha_k \lambda_i^k$$

We are given:

$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & e^{2t}+e^{-t}\end{bmatrix}$$

From this, we know we have a characteristic polynomial of:

$$\lambda^2-\lambda -2 = 0 \implies \lambda_1 = -1, ~\lambda_2 = 2$$

From $(1)$, we have $n = 2$ (the number of eigenvalues) and can write:

$$\tag 3 e^{At} = \alpha_0 ~I + \alpha_1 ~ A$$

From $(2)$, we can write:

$$\alpha_0 - \alpha_1 = e^{-t} \\ \alpha_0 + 2 \alpha_1 = e^{2t}$$

Solving this $~2x2~$ yields:

$$\alpha_0 = \dfrac{1}{3} e^{2t} + \dfrac{2}{3}e^{-t}, ~~ \alpha_1 = \dfrac{1}{3}e^{2t} - \dfrac{1}{3}e^{-t}$$

From $(3)$, we have:

$$A = \dfrac{e^{At}-\alpha_0 I}{\alpha_1}$$

Substituting all of the information, we arrive at:

$$A = \dfrac{1}{2}\begin{bmatrix} 1 & 3 \\ 3 & 1 \end{bmatrix}$$

Of course, we can use this calculated $A$ and determine $e^{At}$, yielding:

$$e^{At} = 1/2\begin{bmatrix}e^{2t}+e^{-t} & e^{2t} - e^{-t} \\ e^{2t}-e^{-t} & e^{2t}+e^{-t}\end{bmatrix}$$

Answer 2

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $$ \left.\mbox{Note that}\ A = \totald{\pars{\expo{At}}}{t}\right\vert_{t\ =\ 0} $$

\begin{align} \color{#00f}{\large A} &= \totald{}{t}\bracks{\half\pars{\begin{array}{cc} \expo{2t} + \expo{-t} & \expo{2t} - \expo{-t} \\[2mm] \expo{2t} - \expo{-t} & \expo{2t} + \expo{-t} \end{array}}}_{t\ =\ 0} = \half\pars{\begin{array}{cc} 2\expo{2t} - \expo{-t} & 2\expo{2t} + \expo{-t} \\[2mm] 2\expo{2t} + \expo{-t} & 2\expo{2t} - \expo{-t} \end{array}}_{t\ =\ 0} \\[3mm]&= \color{#00f}{\large\half\pars{\begin{array}{cc} 1 & 3 \\[2mm] 3 & 1 \end{array}}} \end{align}