Every invertible matrix can be written as the exponential of one other matrix I'm looking for a proof of this claim: "every invertible matrix can be written as the exponential of another matrix". 
I'm not familiar yet with logarithms of matrices, so I wonder if a proof exists, without them. I'll be happy with any proof anyways. I hope someone can help.
 A: As julien has pointed out, presumably we work over the complex field, as the statement is not true over $\mathbb{R}$ (e.g. $e^x=-1$ has no real solution). Then here is an elementary proof for the statement. Let $A=S\left(J_{k_1}(\lambda_1)\oplus\cdots\oplus J_{k_m}(\lambda_m)\right)S^{-1}$ be a Jordan decomposition of $A$, where each eigenvalue is nonzero. Then $\lambda_i=e^{z_i}$ for some $z_i\in\mathbb{C}$. Now consider the Jordan block $Y_i=J_{k_i}(z_i)$. Then $\exp(Y_i)$ is an upper triangular matrix that is always similar to $J_{k_i}(\lambda_i)$. That is, $J_{k_i}(\lambda_i)=P_i\exp(Y_i)P_i^{-1}$ for some invertible matrix $P_i$. Now define $X_i=P_iY_iP_i^{-1}$. Then $A=\exp\left(S(X_1\oplus\ldots\oplus X_m)S^{-1}\right)$.
A: I assume you are talking about complex $n\times n$ matrices. This is not true in general within real square matrices.
A simple proof goes by functional calculus. If $A$ is invertible, you can find a determination of the complex logarithm on some $\mathbb{C}\setminus e^{i\theta_0}[0,+\infty)$ which contains the spectrum of $A$. Then by holomorphic functional calculus, you can define $B:=\log A$ and it satisfies $e^B=A$.
Notes:
1) There is a formula that says $\det e^B=e^{\mbox{trace}\;B}$ (easy proof by Jordan normal form, or by density of diagonalizable matrices). Therefore the range of the exponential over $M_n(\mathbb{C})$ is exactly $GL_n(\mathbb{C})$, the group of invertible matrices.
2) For diagonalizable matrices $A$, it is very easy to find a log. Take $P$ invertible such that $A=PDP^{-1}$ with $D=\mbox{diag}\{\lambda_1,\ldots,\lambda_n\}$. If $A$ is invertible, then every $\lambda_j$ is nonzero so we can find $\mu_j$ such that $\lambda_j=e^{\mu_j}$. Then the matrix $B:=P\mbox{diag}\{\mu_1,\ldots,\mu_n\}P^{-1}$ satisfies $e^B=A$.
3) If $\|A-I_n\|<1$, we can define explicitly a log with the power series of $\log (1+z)$ by setting $\log A:=\log(I_n+(A-I_n))=\sum_{k\geq 1}(-1)^{k+1}(A-I_n)^k/k.$ 
4) For a real matrix $B$, the formula above shows that $\det e^B>0$. So the matrices with a nonpositive determinant don't have a log. The converse is not true in general. A sufficient condition is that $A$ has no negative eigenvalue. For a necesary and sufficient condition, one needs to consider the Jordan decomposition of $A$.
5) And precisely, the Jordan decomposition of $A$ is a concrete way to get a log. Indeed, for a block $A=\lambda I+N=\lambda(I+\lambda^{-1}N)$ with $\lambda\neq 0$ and $N$ nilpotent, take $\mu$ such that $\lambda=e^\mu$ and set $B:=\mu+\log(I+\lambda^{-1}N)=\mu+\sum_{k\geq 1}(-1)^{k+1}\lambda^{-k}N^k$ and note that this series has actually finitely many nonzero terms since $N$ is nilpotent. Do this on each block, and you get your log for the Jordan form of $A$. It only remains to go back to $A$ by similarity.
6) Finally, here are two examples using the above:
$$
\log\left( \matrix{5&1&0\\0&5&1\\0&0&5}\right)=\left(\matrix{\log 5&1&-\frac{1}{2}\\ 0&\log 5&1\\0&0&\log 5} \right)
$$
and 
$$
\log\left(\matrix{-1&0\\0&1} \right)=\left(\matrix{i\pi&0\\0&0} \right)
$$
are two possible choices for the log of these matrices.
A: see this book on the net, it can be helpful
Miroslav Fiedler, Special Matrices and Their Applications in Numerical Mathematics, Springer (1986) 
