$A = B^2$ for which matrix $A$? Is it true that for any $A\in M(n,\mathbb{C})$there exist a $B\in M(n,\mathbb{C})$ such that $A = B^2$? I think this is not true (but I don't know nay example), and then is it possible to characterize such $A$? 
 A: No, it is not true that every complex square (no pun intended) matrix is the square of at least one other such matrix. The simplest counterexample is $\left(\begin{smallmatrix}0&1\\0&0\end{smallmatrix}\right)$.
Here is a rapid analysis. Any square root of $A$ will commute with $A$, so it will stabilise all its generalised eigenspaces (kernels of $(A-\lambda)^k$ for appropriate $\lambda\in\mathbf C$, $k\in\mathbf N$), and therefore the direct sum decomposition of the whole space into such subspaces. This means we can focus on individual generalised eigenspaces, which amounts to assuming that $A$ has only one eigenvalue $\lambda$.
If $A$ has a single eigenvalue $\lambda\neq0$ then we can write $A=\lambda(I+N)$ where $N$ is a nilpotent matrix, and in this case a square root is given by $\sqrt\lambda\sqrt{I+N}$ where $\sqrt\lambda$ denotes one of the complex square roots of $\lambda$, and $\sqrt{I+N}$ is given by Newton's binomial formula
$$
  \sqrt{I+N}=\sum_{k=0}^\infty\binom{1/2}kN^k,
$$
which is actually a finite sum since $N$ is nilpotent.
So we are left with the case that $A$ is itself nilpotent, and here a square root does not always exist. On a suitable basis $A$ will be in Jordan normal form, and whether $A$ has a square root depends on the sizes of the Jordan blocks. A square root will in any case itself be nilpotent, so it is useful to look what happens with the sizes of the Jordan blocks when we take the square. It is easy to see that any Jordan block of size $d>1$ gives upon squaring two Jordan blocks of sizes $\lceil\frac d2\rceil$ and $\lfloor\frac d2\rfloor$. Inversely therefore, a nilpotent matrix has a square root if and only if the sizes of its Jordan blocks can be grouped two at a time, with possibly an isolated block of size $1$, such that two block sizes grouped together never differ by more than$~1$. So the above example which has a single block of size $2$ has no square root, but if we extend it to size $3\times3$ by a row and column of entries $0$ it does:
$$
  \begin{pmatrix}0&1&0\\0&0&0\\0&0&0\end{pmatrix} = 
  \begin{pmatrix}0&0&1\\0&0&0\\0&1&0\end{pmatrix}^2.
$$
A single nilpotent Jordan block of size $3$ however will remain without square root even if we make it arbitrarily large by extending with zeros. In general, the criterion for nilpotent matrices is that between any pair of absent block size values, the number of blocks with intermediate sizes must be even for a square root to exist. For instance for sizes $7,6,4,4,4,3,3,1,1,1$ no square root exists, because between the absent sizes $5$ and $2$ there are five blocks, of sizes $4$ or $3$.
A: The Cholesky decomposition is loosely related to the concept of taking the square root of a matrix. If $A$ is a positive-definite Hermitian matrix, then $$ A = B B^{*} $$, where $B$ is lower triangular with positive diangonal elements where $B^*$ is the conjugate transpose of B.
