Compute $e^A$ where $A$ is a given $2\times 2$ matrix

Question

Compute $e^A$ where $A=\begin{pmatrix} 1 &0\\ 5 & 1\end{pmatrix}$

definition

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

Answer 1

Or, write

$A = I + N \tag{1}$

with

$I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \tag{2}$

and

$N = \begin{bmatrix} 0 & 0 \\ 5 & 0 \end{bmatrix}. \tag{3}$

Note that

$IN = NI = N, \tag{4}$

that is, $N$ and $I$ commute, $[N, I] = 0$, and apply the well-known result that for commuting matrices $B$ and $C$ we have $e^{B + C} = e^B e^C$, an exposition of which may be found here: $M,N\in \Bbb R ^{n\times n}$, show that $e^{(M+N)} = e^{M}e^N$ given $MN=NM$,

and write

$e^A = e^I e^N. \tag{6}$

Now using the matrix power series definition of $e^X$,

$e^X = \sum_0^\infty \dfrac{X^n}{n!}, \tag{7}$

it is easily seen that

$e^I = \sum_0^\infty (\dfrac{1}{n!} I) = (\sum_0^\infty \dfrac {1}{n!}) I = e I, \tag{8}$

since $I^n = I$ for all $n$, while we easily calculate

$N^2 = 0, \tag{9}$

so with $X = N$ (7) becomes

$e^N = I + N = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}. \tag{10}$

Then

$e^A = e^{I + N} = e^I e^N$ $ = e^I(I + N) = e I(I + N) = e (I + N) , \tag{11}$

and finally, again with the aid of (10),

$e^A = \begin{bmatrix} e & 0 \\ 5e & e \end{bmatrix}. \tag{12}$

Answer 2

i will turn it into $$\frac{dx}{dt} = x, \frac{dy}{dt} = 5x + y$$ which has solution $$x = Ae^t, \frac{d}{dt}\left(ye^{-t}\right) = 5A \to y = Be^t + 5At e^t $$

we can write $$\pmatrix{x\\y} = \pmatrix{e^t&0\\5te^t&e^t}\pmatrix{x(0)\\y(0)} $$ by a property of $e^{At}$ this implies $$ e^{tA} = \pmatrix{e^t&0\\5te^t&e^t},\, e^{A} = \pmatrix{e&0\\5e&e} $$

Answer 3

Another way of proving this...

we know from the property of $e$, that if $S,T$ are linear operators and $ST=TS$ then $$e^{S+T}=e^S.e^T$$

Letting $S=I$ and $T=\begin{pmatrix}0&0\\5&0\end{pmatrix}$

Then $$e^{S+T}=e^S.e^T=e^I.e^{\begin{pmatrix}0&0\\5&0\end{pmatrix}}=e(I+T)=e\begin{pmatrix}1&0\\5&1\end{pmatrix}$$

Since, $T^2=0=T^3=\ldots$