For every matrix $A\in M_{2}( \mathbb{C}) $ there's  $X\in M_{2}( \mathbb{C})$ such that $X^2=A$? True\False?
For every matrix $A\in M_{2}( \mathbb{C}) $ there's  $X\in M_{2}( \mathbb{C})$ such that $X^2=A$.
I know that every complexed matrix has a Jordan form matrix $J$ such that $P^{-1}CP=J$, But it's not diagonalizable for sure.
Thanks 
 A: This answer is just to indicate that in Jonas Meyer's answer the second sentence "In general..." is not correct: $$\left(\begin{matrix}0&0&1\\0&0&0\\0&1&0\end{matrix}\right)^2=\left(\begin{matrix}0&1&0\\0&0&0\\0&0&0\end{matrix}\right).$$
I mentioned this as comment, but the matrices won't typeset correctly in a comment.
Now that I'm writing an answer anyway, I'll add that for any other $2\times2$ non-diagonalisable matrix, which one can write as $\lambda(I_2+N)$ with $\lambda\neq0$ and $N^2=0$, one can explicitly get a square root $\sqrt\lambda(I_2+\frac12N)$, where $\sqrt\lambda$ is one of the two complex square roots of$~\lambda$. See this answer for the general case of square roots of complex square $n\times n$ matrices.
A: The $2$-by-$2$ matrices with no square roots are precisely those with Jordan form $\begin{bmatrix}0&1\\0&0\end{bmatrix}$.  In general, every Jordan block corresponding to the eigenvalue $0$ has to have size $1$ in order for a square root to exist.  @Did indicates an elegant way to see this in a comment.  
The Wikipedia article on matrix functions indicates how functions can be applied to Jordan blocks provided sufficiently high order derivatives are defined at the eigenvalues, and this includes square root functions.  For further details, a wonderful reference is Higham's Functions of matrices: theory and computation.  In particular, a sufficient condition for the existence of square roots is that every Jordan block corresponding to the eigenvalue $0$ has size $1$.  
Thanks to Marc for pointing out errors in the old version.  See his answer.
A: As I was recommended , I figured out the answer, with the help of @Did and @JonasMeyer and here is my full explanation.
If A is diagonalizable, there's nothing to prove, cause we are under $\mathbb C$, so $X$ will be simply the roots of A's eigenvalues in the diagonal. 
Let's assume that A is not diagonalizable, we know that every matrix under $\mathbb C$ has a Jordan form, so one of the jordan form that comes to the head is this nilpotent matrix : $A=\begin{pmatrix}
0 &1 \\ 
 0&0 
\end{pmatrix}$ . Let's assume that there's $X$ so that $X^2=A$, and $A^2=0$, so $X^4=0$, so $X$ is nilpotent as well, but it can't be $0$ , so it has to implies $X^2=0$, But $X^2=0\neq A$, Contradiction.
So We find one matrix that does not have such $X$.
