Let $A\in M\_n(\mathbb C)$ be arbitrary. I'm interested to know How should $A^{\alpha}$ be defined for real $\alpha\in [0,\infty)$? When $A$ is nonsingular, we can define $A^{\alpha}=\exp(\alpha \log(A))$ for $\alpha \in [0,1]$ and $A^{\alpha}=A^{[\alpha]}.A^{\alpha-[\alpha]}$ ([.] denote the intiger part) for $\alpha \gt 1$. But what we can do about singular matrices? 



How should $A^α$ be defined for real $α ∈ [0,∞)$ and $A\in M_n(\mathbb C)$?
Let me suggest a solution since I don't think that any of the above comments address the essentials. Firstly, there is no need to use the logarithm to attack this problem. One can, of course, but I think that its use just muddies the water. There is a standard path to address this kind of problem---the Jordan canonical form (see any respectable text on linear algebra). This allows one to reduce to the case where our matrix is a Jordan form, i.e., with identical elements in the diagonal, $1$'s in the superdiagonal and zeroes elsewhere. As pointed out by the OP there is no problem if the diagonal elements are non-zero (we remark that one can then construct the powers directly using the generalised binomial theorem. In this case the resulting infinite series is actually finite since the perturbing matrix is nilpotent). Of course, if the Jordan bock corresponds to an eigenvalue which is not on the positive real axis, in particular if negative, then one then has the task or privilege of deciding on a suitable choice of a branch of the power function, but one already has this problem in the one-dimensional case, i.e. when taking powers of complex numbers (at this point one would use logarithms, of course). The case where the diagonal elements are zero is a train wreck as can be seen from the simplest case where $A$ is the $2 \times 2$ Jordan matrix with zeroes in the diagonal. This does not have a square root, i.e., a $B$ with $B^2=A$ and so things fall apart.


