It's well known (wikipedia: Square root of a matrix) that we can calculate a square root of a matrix by diagonalization or Jordan decomposition (if it's possible). But I saw a different method which seems to work, although it isn't obvious for me why it works.
The method for a diagonalizable matrix $A$:
First calculate the minimal polynomial $\mu_A$ of the matrix. Then chose a polynomial $p$ with $p(\lambda)=\sqrt{\lambda}$ for every eigenvalue $\lambda$.
Now we get $\sqrt{A}=p(A)$.
But does it always work and why? It seems to be possible to use the method in a similar way, if the matrix isn't diagonalisable e.g. if the multiplicity $\lambda$ as a zero of $\mu$ is greater than one. In this case we have to add conditions to the derivates of $p$, like $p'(\lambda)=\frac{1}{2\sqrt{\lambda}}$. Is this also correct?
I think I have an answer to my own question:
We have $p^2(\lambda)=\lambda$ for all eigenvalues of $A$. So by euclidean division we get $p^2=q\cdot \mu_A+r$ where the remainder have the properties $r(\lambda)=\lambda$ and $\deg(r)<\deg(\mu_A)$. So we get $r(t)=t$, because we have $\deg(\mu_A)$ non trivial conditions on $r$.
So we conclude that $(p(A)^2)=p^2(A)=r(A)=A$ and so $p(A)$ is a square root of $A$.
With similar arguments we get the case where $A$ isn't diagonalizable.
But I still wonder why the method is so unknown, when it's a efficient method to calculate square roots of non diagonalizable matrices...
EDIT: I've found some other results now: Why does calculating matrix inverses, roots, etc. using the spectrum of a matrix work?; Wikipedia: Sylvester's formula