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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?

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    $\begingroup$ I don't think you have a choice. $\endgroup$ – Gregory Nov 16 '20 at 16:19
  • $\begingroup$ One other choice might be to calculate via $\exp(M) = I/(0!) + M/(1!) + M^2/(2!) + \cdots$. But that's certainly not an easier way. $\endgroup$ – aschepler Nov 17 '20 at 1:07
  • $\begingroup$ @aschepler Are you sure? $\endgroup$ – Rodrigo de Azevedo Nov 17 '20 at 1:53
  • $\begingroup$ @RodrigodeAzevedo I see your point! $\endgroup$ – aschepler Nov 17 '20 at 17:59
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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 $$e^M = p(M) = \frac{\sin\sqrt{89}}{\sqrt{89}}M+ \cos\sqrt{89}I = \left( \begin{array}{cc} \frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89}& -\frac{(6+2 i) \sin \sqrt{89}}{\sqrt{89}} \\ \frac{(6-2 i) \sin \sqrt{89}}{\sqrt{89}} & -\frac{7 i \sin \sqrt{89}}{\sqrt{89}} +\cos\sqrt{89} \\ \end{array} \right).$$

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  • $\begingroup$ $p$ actually has degree less than 2, which is as it should be. $\endgroup$ – ccorn Nov 16 '20 at 17:38
  • $\begingroup$ @ccorn I remembered it like this: if $m(x) = (x-\lambda_1)^{r_1}\cdots(x-\lambda_k)^{r_k}$ is the minimal polynomial of our matrix $M$, then we need a polynomial $p \in \Bbb{C}[x]$ such that $$p^{(j)}(\lambda_i) = f^{(j)}(\lambda_i), \quad 0 \le j \le r_i-1$$ for all $1 \le i \le k$. Then $p(M) = f(M)$. There is a unique polynomial $p$ of degree $\sum_{i=1}^k r_i = \deg m$ such that this holds. Am I misremembering something? $\endgroup$ – mechanodroid Nov 16 '20 at 17:45
  • $\begingroup$ The calculation in your answer is correct, but the unique second degree must be changed to unique less-than-second degree. Because interpolating $n$ values gives you a unique polynomial of degree less than $n$. Your $\sum_{i=1}^k r_i$ is the degree $n$ of the minimal polynomial, not of the resulting interpolation polynomial. $\endgroup$ – ccorn Nov 16 '20 at 17:59
  • $\begingroup$ @ccorn You seem to be right, but what is the degree of $p$ in general? $\endgroup$ – mechanodroid Nov 16 '20 at 18:03
  • $\begingroup$ That depends on $f$: Extreme example: $f(x)=\text{const}$ always gives a constant polynomial, regardless of matrix size. And $f(x)=x$ always results in a linear polynomial. $\endgroup$ – ccorn Nov 16 '20 at 18:05
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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

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

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    $\begingroup$ This is a wonderful solution. Never seen such a simple yet interesting use of Cayley - Hamilton ! Very nice. $\endgroup$ – Kolmogorov Nov 16 '20 at 19:36
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    $\begingroup$ Agreed, this is very clever. $\endgroup$ – Gregory Nov 19 '20 at 1:19
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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.

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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.

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    $\begingroup$ Pretty, ain't it? $\endgroup$ – Rodrigo de Azevedo Nov 23 '20 at 21:18
  • $\begingroup$ Even prettier: notice that the subalgebra generated by M is isomorphic to complex numbers, this you can apply Euler's identity directly, eliminating some $i$ terms here. $\endgroup$ – lisyarus Nov 28 '20 at 0:01

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