Exponential of a square matrix [closed]

Question

I need to find the matrix exponential $\exp(At)$ where

$$A= \begin{pmatrix} -a & 0 \\ 1 & 0 \\ \end{pmatrix}. $$

Answer 1

one way is to use the power series for $e^{At}$ as suggested by Marcel in the comments.

here is another way to find $e^{at}.$ i will use the property that $e^{At}x_0$ is the unique solution of $$\frac{dx}{dt} = Ax, x(0) = x_0.$$ in this example, we have the decoupled system of two equations:$$\frac{dx}{dt} = -ax, \frac{dy}{dt} = x, x(0) = x_0, y(0) = y_0 $$ the solution is $$x = e^{-at}x_0, y = \frac{x_0}a\left(1- e^{at}\right) + y_0 $$ we can write this as $$\pmatrix{x\\y} = \pmatrix{e^{-at}&0\\\frac{1}a\left(1- e^{at}\right)&1} \pmatrix{x_0\\y_0.}$$ therefore $$e^{At} = \pmatrix{e^{-at}&0\\\frac{1}a\left(1- e^{at}\right)&1} $$

p.s. the exponential growth/decay model $$\frac{dx}{dt} = -ax, x(0)=x_0$$ has the solution $x=e^{-at}x_0.$ we can solve $$ \frac{dy}{dt} = x= e^{-at}x_0, y(0) = y_0$$ has the solution $$y = y_0+x_0\int_0^t e^{-at} \, dt $$ i will let you do the integral to arrive at the solution for $y.$

Answer 2

Another way, that's rather special for $2 \times 2$ matrices $A$: if $A$ has distinct eigenvalues $\lambda_1, \lambda_2$ then

$$ \exp(tA) = e^{\lambda_1 t} \dfrac{A - \lambda_2 I}{\lambda_1 - \lambda_2} - e^{\lambda_2 t} \dfrac{A - \lambda_1 I}{\lambda_1 - \lambda_2}$$

In this case the eigenvalues are $-a$ and $0$, so $$\exp(tA) = \dfrac{1-e^{-at}}{a} A + I = \pmatrix{e^{-at} & 0\cr (1-e^{-at})/a & 1\cr}$$

Answer 3

We have to diagonalize $A$ first and then use that $\exp(\mathrm{diag}(l_1,l_2))=\mathrm{diag}(\exp(l_1),\exp(l_2)))$ at the end use the conjugacy matrices to write down the solution on the original coordinate system.