Existence of square root of a matrix Testing a method with the use of C.-H. theorem for  finding square roots of real $ 2 \times 2 $ matrices I have noticed that some matrices probably don't have  their square roots with  real and complex entries.   
An example the matrix $A= \begin{bmatrix}  0 & 1 \\ 0 & 0 \end{bmatrix}$.
However is it at all a proof that it is impossible  to extend somehow the field of entries in order to satisfy equation $B^2=A$ similar to the situation when many years ago solution of $a^2=-1$ seemed to be impossible to solve for real numbers hence imaginary numbers $i$ were introduced ?
Is it   possible to devise such numbers (...quaternions? octonions ? or others..) that $B^2=A$ would be however satisfied ?
Additionally, when we are sure in general case for $n \times n$ matrices that a square root exists  if we are free to vastly extend a field?
 A: Two partial answers to your question make a full answer!
Let $A=\begin{bmatrix}0&1\\0&0\end{bmatrix}$.


*

*Answer 1: The set of matrices form a ring (rings are sets where you have algebraic operations of addition and multiplication).  In abstract algebra, one learns about ring extensions, in other words, you can construct a bigger ring which contains the ring of matrices, but your given matrix $A$ has a square root.  This is easier to do in a commutative ring, but matrix multiplication is not commutative.
In this case, let $M_{2,\mathbb{R}}$ be the set of $2\times 2$ matrices with coefficients in $\mathbb{R}$, and we consider elements which are sums of matrices times $B$.  In other words, you have sums whose terms are like
$$
BM_1BM_2BM_3B\cdots M_kB
$$
with or without the leading $B$'s.  The one extra condition is that $B^2=A$.
This gets complicated, but does include a square root of $B$.  The problem is that $B$ is not a matrix, it's just an extra element in the ring that acts like the square root $A$.

*Answer 2: Suppose that $B$ must be a matrix (and we're working in a field of characteristic $0$).  Then, we have the situation
$$
\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}a&b\\c&d\end{bmatrix}=\begin{bmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{bmatrix}.
$$
Let's start with the lower left corner.  We have that $c(a+d)=0$, so either $c=0$ or $a=-d$.  


*

*Let's start with the case where $c=0$.  In this case, the matrix on the RHS simplifies to 
$$
\begin{bmatrix}a^2&ab+bd\\0&d^2\end{bmatrix}
$$
Since the upper left and lower right corners are also $0$, $a^2=0$ and $d^2=0$, so $a=0$ and $d=0$.  But this makes the upper right corner $0$ as well, a contradiction.

*Suppose now that $a=-d$, but then the upper right corner is $b(a+d)=0$, which is also not possible.
Therefore, if $xy=0$ implies that $x=0$ or $y=0$, then there is no way to write $A$ as a square of a matrix no matter what field you work with (assuming in our field that $0\not=1$). 
Concluding remark: If you work with matrices over a commutative ring which has zero divisors (so not an integral domain), then it may be possible to find a matrix which is the square root of $A$. 
A: That follows is the continuation of the @Michael Burr 's post.We assume that the entries of $A=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ are in $R$, a non-commutative ring with unity and we want to solve $X^2=A$ where $X=\begin{pmatrix}p&q\\r&s\end{pmatrix}$. 
Necessarily $AX=XA$, that implies $p=s,r=0$. Finally, $X=\begin{pmatrix}p&q\\0&p\end{pmatrix}$ where $p^2=0,pq+qp=1$.
The simplest example: if $R=M_2(K)$ (where $K$ is a field), then a particular square root (there are many other solutions) is obtained for 
$p=\begin{pmatrix}0&1\\0&0\end{pmatrix},q=\begin{pmatrix}x&y\\1&-x\end{pmatrix}$.
