Part (a)
For which 2×2 orthogonal matrices A does
$\large e^A=I+\frac{A^1}{1!}+\frac{A^2}{2!}+…$
converge?
Part(b)
For what A does the series converge to an orthogonal matrix?
My work:
Let A be 2×2 and orthogonal. Then $A^tA = AA^t = I$ and so this implies that A is normal. Over the ground field = $C$, A is then orthogonally / unitarily diagonalizable.
We can write $A=QDQ^*$, where Q is unitary and D is diagonal with the eigenvalues of A on the diagonal. Also, since A is assumed to be orthogonal, then the modulus of each eigenvalue is 1.
Now $$e^A=I+\frac{A^1}{1!}+\frac{A^2}{2!}+…$$
$$ e^A=I+\frac{QDQ^*}{1!}+\frac{QD^2Q^*}{2!}+…$$
$$ e^A= Q(I+\frac{D}{1!}+\frac{D^2}{2!}+…)Q^*$$
$$ e^A= Qe^DQ^*$$
Where $e^D$ is again diagonal.
What can I say from here? I know that online sources such as Wikipedia and Wolfram just state without any proof or extended discussions that the matrix exponential is well-defined and converges for any square matrix. If this is stated as a fact without proof, then it seems a little strange that I am working on a problem statement that asks "for which orthogonal 2×2 matrices A does $e^A$ converge". Is there an important point that I am overlooking? Or can I really just state that the matrix exponential converges for any square matrix A, hence it is well-defined and converges for any 2×2 orthogonal matrix A?
Any suggestions and hints for how to finish part (a) and how to start on part (b) are welcome.
Thanks,
Best Answer
$\quad$ Every matrix has an element of maximal size. $($Obviously, if anything can cause
divergence, it's that one$).~$ Let its absolute value be $M.~$ So let us construct a square
matrix S, whose every single element is $M.~$ Then $S^k=\Big(n^{k-1}M^k\Big)_{n\times n}~,~$ and each
element of $A^k$ lies between $\pm~n^{k-1}M^k.~$ But $~e^S\approx\bigg(\dfrac{e^{nM}}n\bigg)_{n\times n}~,~$ so every element
of $e^A$ is definitely bounded. However, even in this case divergence could still theoret–
ically happen, if at least one such element $($not necessarily the same$)$ were to freely
oscillate inside a given range, without actually converging to any particular value
within that interval. But this is not possible, since each new term of the infinite series
decreases at an exponential rate, being trapped between $\pm~\dfrac{n^{k-1}M^k}{k!}.$