Deriving Rotation Matrix from Strang pg.230

[interested_learner]

This is another attempt at an earlier question that maybe wasn't done in a readable format.

This was in Strang page 230. Could someone derive this?

<br /> if K = \left[\begin{array} {cc} 0&amp;-1\\1&amp;0\end{array}\right]<br /> <br /> then \quad e^{Kt} = \left[\begin{array} {cc} cos t &amp; -sin t\\sin t &amp; cos t\end{array}\right]<br />

 

[matt grime]

By working out what exp{Kt} looks like: K has a nice form, so do the powers of K, hence it is possible to simply work out what the sum defining exp{Kt} is.

 

[HallsofIvy]

The eigenvalues of K are i and -i so, over the complex numbers, K is equivalent to the diagonal matrix with i and -i on the diagonal. In fact, since the corresponding eigenvectors are multiples of [1, -i] and [1, i] respectively, we have:
\left[\begin{array}{cc}0 &amp; -1 \\ 1 &amp; 0\end{array}\right]=\left[\begin{array}{cc}1 &amp; 1 \\-i &amp; i \end{array}\right] \left[\begin{array}{cc}i &amp; 0 \\0 &amp; -i\end{array}\right]\left[\begin{array}{cc}\frac{i}{2} &amp; -\frac{1}{2} \\-\frac{i}{2} &amp; -\frac{1}{2}\end{array}\right]

The point of "diagonalizing" like that is that if A= CDC^{-1} then A^2= (CDC^{-1})(CDC^{-1})= CD^2C^{-1} and similarly for higher powers. Then, using the Taylor's series for e^x:e^x= 1+ x+ \frac{1}{2}x^2+ \cdot\cdot\cdot + \frac{1}{n!}x^n+ \cdot\cdot\cdot, we have
e^{CDC^{-1}}= I+ CDC^{-1}+ \frac{1}{2}CD^2C^{-1}+ \cdot\cdot\cdot+ \frac{1}{n!}CD^nC^{-1}+ \cdot\cdot\cdot
Since I= CC^-1, that is
C(I+ D+ \frac{1}{2}D^2+ \cdot\cdot\cdot + \frac{1}{n!}D^n+ \cdot\cdot\cdot = Ce^DC^{-1}
For a diagonal matrix, powers just give powers on the diagonal and adding just adds the diagonal values, for diagonal matrix D, e^D is just the diagonal matrix with exponentials on the diagonal. In this case
e^{Kt}= \left[\begin{array}{cc}1 &amp; 1 \\-i &amp; i \end{array}\right]\left[ \begin{array}{cc}e^{it} &amp; 0 \\ 0 &amp; e^{-it}\end{array}\right]\left[\begin{array}{cc}\frac{i}{2} &amp; -\frac{1}{2} \\-\frac{i}{2} &amp; -\frac{1}{2}\end{array}\right]
Use e^{it}= cos(t)+ isin(t) and multiply it out.

 

[interested_learner]

Wow! Thanks

Wow! Such a quick and clear answer. Thanks!