find $\alpha,\beta$ such that $\exp(A)=\alpha(\omega)\mathbb{I}_{2}+\beta(\omega)A$

Question

I have a matrix $A =\begin{bmatrix}a & b\\c & -a\end{bmatrix}$ where $a,b,c\in\mathbb{R}$

Now, I have to show that I can write $\exp(A)$ in a way $\alpha(\omega)\mathbb{I}_2+\beta({\omega})A$, such that both functions $\alpha,\beta$ only depend on det$(A)$.

As a hint, I was supposed to distinguish between three cases: det$(A)$ =,<,> $0$ and use series representations of trigonometric and hyperbolic functions.

After calculating $A^2$ case det$(A)=0$ is obvious, however I'm a bit stuck on how to proceed for the other two cases, where I mainly can't really see how to invoke trigonometric and hyperbolic functions.

Answer 1

Since $$e^A=\sum_{k=0}^\infty\frac{1}{k!}A^k,$$ and $$A^2=-det(A)I_2\quad \text{(verify!)}.$$ We conclude $$e^A=\left(\sum_{k=0}^\infty\frac{(-det\ A )^k}{(2k)!}\right)I_2+\left(\sum_{k=0}^\infty\frac{(-det\ A )^k}{(2k+1)!}\right)A.$$

Answer 2

More generally, for any analytic function $f$ and any $2 \times 2$ matrix $A$ we have $f(A) = \alpha I + \beta A$ where:

1. If $A$ has two distinct eigenvalues $\lambda_1$ and $\lambda_2$, then $f(\lambda_1) = \alpha + \beta \lambda_1$ and $f(\lambda_2) = \alpha + \beta\lambda_2$.
2. If $A$ has just one eigenvalue $\lambda$, then $f(\lambda) = \alpha + \beta \lambda$ and $f'(\lambda) = \beta$.

Answer 3

$\newcommand{I}{\mathbb{I}_2}$ Okay, I'm probably making a mistake, since I'm not using the hint, but here's what I've got!

First of all, by definition of the matrix exponential, we have $$ \exp(A):=\sum_{k=0}^\infty\frac{1}{k!}A^k = \I + A + \frac{1}{2}A^2 + \dots $$ Simple calculations shows that $$ |A|:=\det(A)=-a^2-bc, \\ A^2 = -|A|\;\I. $$ and therefore by induction $$ A^k= \begin{cases} (A^2)^{k/2}&=(-|A|)^{k/2}\;\I,&\quad\text{$k$ even},\\ (A^2)^{(k-1)/2}A&=(-|A|)^{(k-1)/2}\;A,&\quad\text{$k$ odd}. \end{cases} $$ Substituting this into the definition of the matrix exponential gives $$ \exp(A) = \sum_\text{$k$ even}\left[\frac{1}{k!}(-|A|)^{k/2}\right]\I + \sum_\text{$k$ odd}\left[\frac{1}{k!}(-|A|)^{(k-1)/2}\right]A = \alpha\;\I+\beta A, $$ where $\alpha, \beta$ are real numbers that only depend on $|A|$.