Can we take the square root of both sides of a matrix equation? Let $\mathbf{A}$ and $\mathbf{B}$ two square matrices of size $n$ and $\mathbf{A^2=B^2}$. 
When can we say that $\mathbf{A=B}$. 
Essentially, the question is: When does a matrix have a unique square root? 
Now, I know that positive (semi) definite matrices have this property, but, can we claim the converse? 
I.e., does a matrix have unique square root if and only if is psd?
Thank you.
Edit:
Firstly, let my clarify that the matrices are in $\mathbb C^{n\times n}$.
From what I gathered from your replies, there is never uniqueness of the square root except in the case of the all-zero matrix of size $n=1$. 
Moreover, there can be infinite roots even for psd matrices.
From other sources I know that a psd will have only one psd root.
Now let me pose a similar but distinctly different question. Is defectiveness a criterion for existence of root?
 A: A reasonable class of matrices where uniqueness of square root holds, is the class of all hermitian positive semidefinite matrices over $\mathbb{C}$. However, you need to be looking for the square root in this class, otherwise there is no uniqueness. Without positive semidefiniteness, of course, there are $2^n$ square roots of $\mathrm{diag}(\lambda_1,\ldots, \lambda_n)$, as long as all $\lambda$'s are nonzero (and there might be infinitely many which are not hermitian, as people demonstrated above).
If you consider the class of all matrices over complex numbers, then I believe that only the zero matrix $1\times 1$ has a unique square root; that is, only one single matrix in one single dimension.
A: If $A\neq 0$ and $a^2 = 0$, as soon as you are on a field with characteristic $\neq 2$ then
$$
(-a)^2 = A
$$as well. Then, the only matrix that has only one square root is $0$; it works only in dimension 1 as
$$
\begin{pmatrix}
  0 & 1 \\
  0 & 0
 \end{pmatrix}^2 = 0
$$

Note that there are matrices who do not have square roots at all, such as
$$
\begin{pmatrix}
  0 & 1 \\
  0 & 0
 \end{pmatrix}
$$
