For which orthogonal matrices does the matrix exponential converge?

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 2x2 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 2x2 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 2x2 orthogonal matrix A?

Any suggestions and hints for how to finish part (a) and how to start on part (b) are welcome.

Thanks,

• Your calculation is sufficient for the case you're given. You don't need to appeal to a general result, even though it is nice. What does your calculation say? Oct 13, 2015 at 3:36
• That's exactly right. Oct 13, 2015 at 3:52
• Is it allowed to use property that $e^{A} \cdot e^{B} = e^{B} \cdot e^{A} = e^{A+B} \iff \lbrack A, B \rbrack = 0$? Oct 13, 2015 at 4:24
• Yes, I thought about part (b) when I was asking this. It's easy to see that $(e^{A})^{T} = e^{A^T}$. You want $e^A$ to be orthogonal, i.e. $e^A \cdot e^{A^T} = e^{A^T} \cdot e^A = I$. From the property follows that $e^A \cdot e^{A^T} = e^{A+A^T}$. Now you can conclude what matrix should be $A + A^{T}$ if it's exponent is $I$ :) Oct 13, 2015 at 4:51
• However, take my advice with caution. I need to check is the property that I'm referring to is really true. Oct 13, 2015 at 5:16

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.

$\quad$ 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 $\pm~\dfrac{n^{k-1}M^k}{k!}.$

• Hi @Lucian, can you please elaborate on how you arrive at $n^{k-1}M^k$ in your first equation? And how does the approximation of $e^S$ show the boundedness of $e^A$? Thanks, Oct 14, 2015 at 1:23
• @LebronJames: Just take a simple $2\times2$ or $3\times3$ matrix, and square it. What do you get in each cell ? $M\cdot M+M\cdot M+\ldots+M\cdot M$, n terms in total, meaning $n\cdot M^2$. Multiply it again by itself, and see what happens. Also, it goes on without saying, that if some terms were smaller than M, and positive, then the sum of products would be smaller than $n\cdot M^2$. And if they are negative, and less than M in absolute value, than it's even “worse”, since they not only add less, but outright cancel the contribution of the other positive terms. Oct 14, 2015 at 2:27
• Ok, got it -- thanks so much @Lucian :-) Oct 14, 2015 at 2:57

The fact that the series for $e^A$ is easy to prove using matrix norms. In particular, you should try to prove the inequality $$\left\|e^A \right\| \leq e^{\|A\|}$$ if $\|\cdot\|$ is a norm (such as the Frobenius norm) which satisfies $\|AB\| \leq \|A\| \cdot \|B\|$ for all matrices $A,B$.

It is, in fact, a well known result that a matrix $A$ will be such that $e^{tA}$ is orthogonal for every $t \in \Bbb R$ if and only if $A$ is skew-symmetric, which is to say that $A^T = -A$. This is a commonly used result in the context of Lie Groups and Lie Algebras.

At the very least, you should try to prove that if $A^T = -A$, then $e^A$ is orthogonal.

I am not sure whether there are any other matrices $A$ for which $e^A$ is orthogonal.

A matrix $A$ such that $e^A$ is orthogonal but $A\neq A^T$: $$\pmatrix{0&1\\0&2\pi i}$$ or better yet $$\pmatrix{0&-1\\ 4 \pi^2 & 0}$$

• @LebronJames Orthogonal matrices do commute with their transpose, because a matrix is orthogonal iff its orthogonal is: $AA^\top=I=A^\top A$. Oct 14, 2015 at 2:53
• I think it's an interesting question. Jordan normal form doesn't help, but Schur decomposition might. Oct 14, 2015 at 3:27
• See my latest edit. Still not sure how to classify all such matrices, though Oct 14, 2015 at 4:10
• Slightly more generally, if $a \ne c$, $$\exp\pmatrix{a & b\cr 0 & c\cr} = \pmatrix{e^a & \dfrac{b (e^c - e^a)}{c-a}\cr 0 & e^c}$$ so this gives an example of a non-normal matrix whose exponential is normal if $a \ne c$ but $e^a = e^c$, i.e. $c - a = 2 \pi i n$ for some nonzero integer $n$. Oct 14, 2015 at 4:31
• For a real example where $e^{A} e^{A^T} = e^{A^T} e^A$ but $e^A e^{A^T} \ne e^{A+A^T}$, try $$A = \pmatrix{0 & 0 & 1 & 0\cr 0 & 0 & 0 & 1\cr 0 & 0 & 0 & -2\pi\cr 0 & 0 & 2\pi & 0\cr}$$ Oct 14, 2015 at 4:41