Why does a positive definite matrix with a repeated eigenvalue have infinitely many square roots? So, if we consider a positive definite matrix $A$, (meaning that $A$ is self-adjoint $(Ax,x) > 0$ and also that $A$ has strictly positive eigenvalues) we see right away that since it is self adjoint, and has an orthonormal basis of eigenvectors $h_{i}$, that it has a uniquely determined positive square root.  We see this, since for the basis
\begin{align}
\{h_{1},...,h_{n}\}
\end{align}
each $x$ can be written as
\begin{align}
x &= \sum_{i = 1}^{n}c_{i}h_{i} \\
\Rightarrow Hx &= \sum_{i=1}^{n}c_{i}Hh_{i} = \sum_{i=1}^{n}c_{i}\lambda_{i}h_{i}.
\end{align}
We let $\sqrt{H}x$ be represented by
\begin{align}
 \sum_{i=1}^{n}c_{i}\sqrt{\lambda_{i}}h_{i}.
\end{align}
where we take $\sqrt{\lambda_{i}}$ to be the positive square root of $\lambda_{i}$. Naturally, one sees that this is the unique positive definite square root, and that there are $2^{n}$ possible square roots, however, I keep reading that if $A$ has a repeated eigenvalue there are infinitely many square roots.  Why is this? 
Please note by square root I mean $S$ such that $S^{2} = A$.

In case you mean "square root" as $C C^T = A:$  
Suppose $A = \lambda^2 I$ two by two,  
$$ C \; = \;  \lambda
 \left(  \begin{array}{rr}
  \cos \theta  &  \sin \theta  \\
   -\sin \theta   &  \cos \theta  
\end{array} 
  \right)  ,
  $$
which is to say that a rotated basis of a two-dimensional eigenspace is being used. 
