Exponential map on matrices I was thinking about different definition of exponential:
Case 1:
Let $z=a+ib$ be a complex number so we have the exponential map $\exp(z)=\sum \frac{z^n}{n!}$.
But one can also view $z$ as a matrix (indeed $z$ is a special automorphism of $\mathbb{R}^2$) of the form $z_r:=Z:=\begin{pmatrix}a & -b \\ b & a \end{pmatrix}$ so one has the exponential matrix map defined by $\exp(Z)=\sum \frac{Z^n}{n!}$ it is not difficult to show that $\exp(z_r)=\exp(z)_r$ (I can give more details for this if it worth, the idea is to use the polar expression of $z$ and that write $Z=a\operatorname{Id}+bI$ and use the Taylor expansion of $\sin$ and $\cos$).
Case 2: (The question)
What about the more general case of the exp function on a linear operator $\phi:\mathbb{C}^n\longrightarrow\mathbb{C}^n$?
Indeed, let $X$ be a complex matrix of size $n$, and let $X_r$ the block-matrix obtained by replacing to each $x_{ij}$ the $2\times2$ matrix ${x{_{ij}}}_r$ is it true that $\exp(X_r)=\exp(X)_r$?
Is it a general thing?
My approach:
I've tried with a calculation like the Case 1 but it seems to be more difficult to calculate. Another approach I have in mind is to use the fact that $\mathbb{C}^n$ is a real smooth variety with standard complex structure and than use the flow theorem. This approach gives me a characterization for $\exp(X_r)$ but my problem is to prove that $\exp(X)_r$ is also a solution for the flow.
(Maybe if this is not very clear I would clarify)
Questions:

*

*Is it possible to prove that $\exp(X_r)=\exp(X)_r$ by straightforward calculation?

*Is it possible to prove that $\exp(X_r)=\exp(X)_r$ by using the flow theorem?

Any hint or suggestion would be great!
 A: Let's dig a bit further into your statement "let $X_r$ the block-matrix obtained by replacing to each $x_{ij}$ the $2\times2$ matrix $(x{_{ij}})_r$". I'd like to reframe that as follows: let $\Phi:M_{n \times n}(\Bbb C) \to M_{2n \times 2n}(\Bbb R)$ denote the map defined by
$$
\Phi(A + Bi) = A \otimes I + B \otimes J,
$$
where $A,B$ are real $n\times n$ matrices, $\otimes$ denotes the Kronecker product, $I$ denotes the size $2$ identity matrix, and
$$
J = \pmatrix{0&-1\\1&0}.
$$
I claim that if $X = A + Bi$, then $X_r = \Phi(X)$. Moreover, I claim the following:

*

*For real $p,q$ and $X,Y \in M_{n \times n}(\Bbb C)$, $\Phi(pX + qY) = p\Phi(X) + q\Phi(Y)$

*For $X,Y \in M_{n \times n}(\Bbb C)$, $\Phi(XY) = \Phi(X)\Phi(Y)$

*For a convergent sequence $X_n \to X$, we have $\Phi(X_n) \to \Phi(X)$. That is, $\Phi(\lim_{n \to \infty}X_n) = \lim_{n \to \infty}\Phi(X_n)$.

These results can be proved using the properties of the Kronecker product and the fact that $J^2 = -I$. Now for any $X \in M_{n \times n}(\Bbb C)$, denote
$$
\exp_N(X) = \sum_{k=0}^N \frac{X^k}{k!}.
$$
Using properties 1 and 2 above, verify that $\Phi(\exp_N(X)) = \exp_N(\Phi(X))$. From there, we can conclude that
$$
\Phi(\exp(X)) = \Phi\left(\lim_{N \to \infty} \exp_N(X)\right)
= \lim_{N \to \infty}\Phi(\exp_N(X)) = \lim_{N \to \infty} \exp_N(\Phi(X))
= \exp(\Phi(X)).
$$
That is, we indeed have $\Phi(\exp(X)) = \exp(\Phi(X))$, which is what we wanted.
A: To elaborate on my comment:

*

*If $X,Y$ commute then $e^{X+Y}=e^X e^Y$. This in particular holds for $X=a\text{ Id}$ and $Y=bI$.


*If $I^2=-\text{Id}$, then $\exp(b I)=(\cos b)\operatorname{Id}+(\sin b)I.$
Both of these may be shown by expanding out the appropriate Taylor series. Combining these, we indeed have
\begin{align}
\exp(z_r)
&=\exp(a\operatorname{Id}+bI)\\
&=\exp(a\operatorname{Id})\exp(bI)\\
&=[(e^a\cos b)\operatorname{Id}+(e^b \sin b)I]\\
&=(e^a\cos b+i e^a\sin b)_r \\
&= (e^{a+ib})_r\\
&= \exp(z)_r.
\end{align}
The limitation of this approach compared to Ben's is that theirs allows matrices $A,B$ whereas mine is specifically for scalar $a,b$.
