# for arbitrary vector $v,u$, is there the matrix X which satisfy the relation exp$[X]\,v=u$?

Nowadays, I'm studying for exponential map of Lie group.

my question is, To make the form of exp$$\begin{pmatrix}x_{11}&x_{12}&\cdots \\x_{21}&\ddots \\ \vdots\end{pmatrix}$$,

I have to multiply same matrixes. so I wonder more basically, when the matrix $$A$$ which has all component can be arbitrary number, although we make $$A^N$$,whether $$A^N$$ has also all component which can be arbitrary number or not.

additionally, My major is physics, by the way, I almost don't know about mathematic symbols. If you know the process of proof, please show me the proof without mathematic symbols possible.

• Its not clear to me what you are asking. What do you mean by "$A^N$ also has all component which can be arbitrary or not"? The components of $A^N$ clearly depend on what the components of $A$ are. – Spencer Apr 23 at 6:51
• Also how is the actual content of your question related to the question title? – Spencer Apr 23 at 6:51
• because the exponential map of $GL(n,\mathcal{C})$ is made by \lim_{N\to \infty}[exp$\begin{pmatrix} x_{11}/N&x_{12}/N&\cdots \\ x_{21}/N&\ddots \end{pmatrix}$]^N. so, I think most general exponential map of Lie group can make all vector for fixed vector which is multiplied to exponential map. – 정재훈 Apr 23 at 7:00
• Thank you for clarifying your question. – Spencer Apr 23 at 7:05
• me too!! thank you for your interest!! – 정재훈 Apr 23 at 7:09