TITLE: For every matrix $A\in M_{2}( \mathbb{C}) $ there's  $X\in M_{2}( \mathbb{C})$ such that $X^2=A$?
QUESTION [22 upvotes]: 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

REPLY [3 votes]: 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.