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I want to calculate $e^{A}$ of the matrix $A$:

$$\left ( \begin{array}{cc} \cos t & \sin t \\ -\sin t & \cos t \end{array} \right )$$

I tried to use $e^{At}=P\ \mbox{diag}(e^{\lambda t}) P^{-1}$, but from there I obtain the eigenvalue as $\cos t-|-\sin t|$ which seemed wrong for the solution of the problem.

Also I tried to use power series, however I couldn't get anything that makes sense.

I don't know what to do, and kind of lost. Could you please help me?

AlexR
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Xentius
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  • Note that $A^n=\left ( \begin{array}{cc} \cos nt & \sin nt \ -\sin nt & \cos nt \end{array} \right )$ – Kitegi May 30 '15 at 21:19

3 Answers3

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The eigenvalues are $\cos(t)+i\sin(t)$ and $\cos(t)-i\sin(t)$. Then what?

GEdgar
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  • Ops, I was calculating the eigenvalues wrong, you are right now everything is solved. Thank you so much! – Xentius May 30 '15 at 21:29
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this is not a complete answer. i need to think about how to sum the cosine and sine series.

the matrix $A$ represents a rotation by an able $t,$ therefore $$A^k = \pmatrix{\cos kt&\sin kt\\-\sin kt & \cos kt}, k \text{ an integer. }$$

now, $$\begin{align}e^A &= I + \frac1{1!}A + \frac1{2!}A^2 + \frac1{3!}A^3 + \cdots \\ &=\pmatrix{1+\frac1{1!}\cos t+ \frac1{2!}\cos 2t+ \cdots &\frac1{1!}\sin t+\frac1{2!}\sin 2t + \cdots\\-\frac1{1!}\sin t - \frac1{2!}\sin 2t -\cdots&1+\frac1{1!}\cos t + \frac1{2!}\cos 2t + \cdots}\\ \end{align}$$

at the suggestion of user farnight, we will look at the expression $$\begin{align}e^{e^{it}} &= e^{\cos t + i\sin t} = e^{\cos t}\left(\cos(\sin t) + i\sin(\sin t)\right)\\ &= 1 + \frac{e^{it}}{1!} + \frac{e^{2it}}{2!} \end{align} $$

taking the real and imaginer parts, we get $$1+\frac1{1!}\cos t+ \frac1{2!}\cos 2t+ \cdots = e^{\cos t}\cos(\sin t) \\ \frac1{1!}\sin t+\frac1{2!}\sin 2t + \cdots= e^{\cos t}\sin(\sin t) $$

therefore $$e^A = \pmatrix{e^{\cos t}\cos(\sin t)& e^{\cos t}\sin(\sin t) \\-e^{\cos t}\sin(\sin t)&e^{\cos t}\cos(\sin t) } $$

abel
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As Farnight points out, $$ A^n = \begin{pmatrix} \cos{nt} & \sin{nt} \\ -\sin{nt} & \cos{nt} \end{pmatrix}, $$ so the exponential sum is directly $$ e^A = \sum_{n=0}^{\infty} \frac{A^n}{n!}, $$ so we have to compute $$ \sum_{n=0}^{\infty} \frac{1}{n!}\cos{nt} \quad \text{and} \quad \sum_{n=0}^{\infty} \frac{1}{n!}\sin{nt} $$ This is easy enough: $$ \sum_{n=0}^{\infty} \frac{1}{n!}\cos{nt} = \frac{1}{2}\sum_{n=0}^{\infty} \left( \frac{e^{int}}{n!} + \frac{e^{-int}}{n!} \right) = \frac{1}{2} (e^{e^{it}}+e^{e^{-it}}) = e^{\cos{t}} \cos{\sin{t}}, $$ which doesn't simplify, and $$ \sum_{n=0}^{\infty} \frac{1}{n!}\sin{nt} = \frac{1}{2i}\sum_{n=0}^{\infty} \left( \frac{e^{int}}{n!} - \frac{e^{-int}}{n!} \right) = \frac{1}{2i} (e^{e^{it}}-e^{e^{-it}}) = e^{\cos{t}}\sin{\sin{t}}, $$ and you can go from there.

(Oh, and note that this works for any complex $t$, whereas some other methods require real $t$, and then a separate generalisation)

Chappers
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