Good evening everyone , I need help on this one please :
Let $a$ be a complex number
and let J defined as the matrix containing the blocks of Jordan matrices
Let $M\in M_n\left(\mathbb{C}\right)$ defined as:
$M=\sum_{k=1}^{n} \frac{(-1)^{k-1}a^{k}J^{k}}{k}$
prove that $exp(M)=I_{n}+aJ$
A few approaches, sketched:
Note that $$ f(x) = \sum_{i=1}^\infty \frac{(-1)^k}{k}x^k $$ is the MacLaurin series for $\ln(1+x)$. In particular: whenever the series converges (when $|x|<1$), $f(x) = \ln(1+x)$ for any scalar $x$. Our matrix $M$ is given by $M = f(aJ)$.
It's easy to see that whenever $J$ is diagonalizable (and $a$ is sufficiently small), $\exp(J) = \exp(\ln(I + aJ)) = I + aJ$. From there, an easy approach is to simply state that this applies to all matrices, since the functions involved are continuous and the diagonalizable matrices are dense.
Alternatively, it suffices to note the composition of power series is another power series, and that with $g(x) = \sum \frac{x^n}{n!} = \exp(x)$. We find that $g(f(x)) = 1 + x$ in the scalar case, which means the same applies in the matrix case.
However, it is unlikely (given the form of the question) that they're looking for an argument from continuity, or via power series alone. So, here's a direct proof in the case of a single Jordan block. In particular, take $J$ to be the $n \times n$ matrix $$ J = \pmatrix{\lambda&1\\&\lambda&\ddots\\&&\ddots\\&&&&1\\&&&&\lambda} $$ We compute $$ M = f(aJ) = \pmatrix{f(a\lambda)&\frac 1{1!}f'(a\lambda)&\frac 1{2!}f''(a\lambda)&\cdots\\ & f(a\lambda)&\frac{1}{1!}f'(a\lambda)&\frac 1{2!}f''(a\lambda)\\ &&\ddots} $$ as described here. Not sure where one would go from there.