Let $C\in\mathbb{R}^{n\times n}$ be a matrix such $det(C)\neq0$ b. I need to prove that exists $B\in\mathbb{R}^{n\times n}$ such $$e^B=C^2$$
This is a question from a curse of ODE, we have seen exponential of a matrix, Jordan form, and I believe this question can be solved with the inverse application theorem since $A\mapsto e^A$ is $C^\infty$ but my work doesn't help me so much. Is there any linear algebra trick or what is the best way to solve this ?
We can find a matrix similar to $C$ in the Jordan canonical form, because $A \exp(B) A^{-1} = \exp(ABA^{-1})$ without loss of generality we can assume that $C$ is in the Jordan canonical form.
It is enough to solve the problem for one Jordan block $J(\lambda)=\lambda I +N$, here $\lambda \neq 0$ ($\det C \neq 0$) and $N$ is nilpotent.
So, we need to solve $e^B=\lambda I + N$. Now we can use series for logarithm $$ \log(1+x) = \sum_{n \geq 1} (-1)^{n-1}\frac{x^n}{n}. $$ Applying it to our case we get $$ B = \log (\lambda I +N)=\log(\lambda) \log(I+\frac{1}{\lambda}N) $$
Matrix $\frac{1}{\lambda}N$ is nilpotent, so $B=\log(\lambda)\log(I+\frac{1}{\lambda}N)$ is a polynomial.
Such matrix $B$ commutes with its complex conjugate $\overline{B}$. Also $\overline{e^B}=e^\overline{B}$, and $\overline{e^B} = \overline{C}=C$. Therefore $e^{B+\overline{B}}=e^B e^\overline{B}=C^2$.