Calculating matrix exponential

Given matrix $$M = \begin{pmatrix} 7i& -6-2i\\6-2i&-7i\end{pmatrix}$$ how do I calculate matrix exponential $e^M$?

I know I can use that $e^A=Pe^DP^{-1}$ where $D=P^{-1}AP$. I computed the characteristic polynomial of the above matrix as

$$P(\lambda)=\lambda^2+89$$

Is there an easier way to do this than trying to compute the diagonalized matrix?

Solution 1:

Via Cayley-Hamilton, ${\rm M}^2 + 89 \, {\rm I}_2 = {\rm O}_2$. Hence,

$$\begin{aligned} {\rm M}^2 &= - 89 \, {\rm I}_2\\ {\rm M}^3 &= - 89 \, {\rm M}\\ {\rm M}^4 &= 89^2 {\rm I}_2\\ {\rm M}^5 &= 89^2 {\rm M}\\ &\vdots\\ {\rm M}^{2k} &= (-1)^k 89^k {\rm I}_2\\ {\rm M}^{2k+1} &= (-1)^k 89^k {\rm M} \end{aligned}$$

and

$$\exp({\rm M}) = \sum_{k=0}^{\infty} \frac{{\rm M}^k}{k!} = \cdots = \color{blue}{\cos( \sqrt{89} ) \, {\rm I}_2 +\frac{\sin( \sqrt{89} )}{\sqrt{89}} {\rm M}}$$

Solution 2:

Your matrix $M$ is diagonalizable with eigenvalues $\pm i\sqrt{89}$. This means that $e^M = p(M)$ where $p \in \Bbb{C}[x]$ is the unique polynomial of degree less than $2$ such that $$p(i\sqrt{89}) = e^{i\sqrt{89}}, \quad p(-i\sqrt{89}) = e^{-i\sqrt{89}}.$$

Using Lagrange interpolation formula, we see that $$p(x) = \frac{x+i\sqrt{89}}{2i\sqrt{89}}e^{i\sqrt{89}}-\frac{x-i\sqrt{89}}{2i\sqrt{89}}e^{-i\sqrt{89}} = \frac{\sin\sqrt{89}}{\sqrt{89}}x+ \cos\sqrt{89}$$ so $$e^M = p(M) = \frac{\sin\sqrt{89}}{\sqrt{89}}M+ \cos\sqrt{89}I = \left( \begin{array}{cc} \frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89}& -\frac{(6+2 i) \sin \sqrt{89}}{\sqrt{89}} \\ \frac{(6-2 i) \sin \sqrt{89}}{\sqrt{89}} & -\frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89} \\ \end{array} \right).$$

Solution 3:

Via Cayley-Hamilton,

$${\rm M}^2 = - 89 \, {\rm I}_2 = \left( i \sqrt{89} \right)^2 {\rm I}_2$$

Hence, matrix ${\rm A} := \frac{{\rm M}}{i \sqrt{89}}$ is involutory, i.e., ${\rm A}^2 = {\rm I}_2$. Using Euler's formula,

$$\begin{aligned} \exp({\rm M}) = \exp \left( i \sqrt{89} {\rm A} \right) &= \cos \left( \sqrt{89} {\rm A} \right) + i \sin \left( \sqrt{89} {\rm A} \right)\\ &= \cos \left( \sqrt{89} \right) {\rm I}_2 + i \sin \left( \sqrt{89} \right) {\rm A}\\ &= \color{blue}{\cos \left( \sqrt{89} \right) {\rm I}_2 + \frac{\sin \left( \sqrt{89} \right)}{\sqrt{89}} {\rm M}}\end{aligned}$$

where $\cos \left( \sqrt{89} {\rm A} \right) = \cos \left( \sqrt{89} \right) {\rm I}_2$ and $\sin \left( \sqrt{89} {\rm A} \right)= \sin \left( \sqrt{89} \right) {\rm A}$ because ${\rm A}$ is involutory.

Solution 4:

You can also use Euclidean divisions to compute the exponential once you have the characteristic polynomial :

$$\forall n\in\mathbb{N}^*,\exists(P_n,R_n)\in\mathbb{R}[X], \deg(R)\leq1, X^n=P_n(X)\times(X^2+89)+R_n(X) $$

Let $R_n(X)=a_nX+b_n$, you get : $$M^n=a_nM+b_n1 $$

and you get the values of $a_n,b_n$ with : $$(i\sqrt{89})^n=a_ni\sqrt{89}+b_n\\ (-i\sqrt{89})^n=-a_ni\sqrt{89}+b_n$$

So you get : $$b_n=\frac{(i\sqrt{89})^n+(-i\sqrt{89})^n}{2}\\ a_n=\frac{(i\sqrt{89})^n-(-i\sqrt{89})^n}{2i\sqrt{89}} $$

And then you use the definition : $$e^M=\sum_{n=0}^{\infty}\frac{M^n}{n!}=\sum_{n=0}^{\infty}\frac{a_nM+b_n1}{n!} $$

And then you can conclude from here.