Confusion about the uniqueness of the square root of a positive definite matrix There is a lot of material showing that a positive definite matrix $A$ has a unique positive definite square root, $B$ such that $B^2=A$. During a self study session, I needed to use this fact for a proof sketch, but I was confused about whether $B$ is the  only square root of $A$ and it happens to be positive definite or is the uniqueness only valid for the positive definiteness; such that one can come up with a matrix $C$, not positive definite and $C^2=A$. I am confused about which is the correct one.
 A: If I think in the case $A=I\in \mathbb{R}^{n\times n}$ the I have clearly at least two posibilities for $B$ and more if $n>1$ for example if $n=2$, there are this posivilities for $B$.
$ \left[ {\begin{array}{cc}
   1 & 0 \\
   0 & 1 \\
  \end{array} } \right]$ $\left[ {\begin{array}{cc}
   1 & 0 \\
   0 & -1 \\
  \end{array} } \right]$ $\left[ {\begin{array}{cc}
   -1 & 0 \\
   0 & 1 \\
  \end{array} } \right]$ or $
 \left[ {\begin{array}{cc}
   -1 & 0 \\
   0 & -1 \\
  \end{array} } \right]
$
but only one of them is positive definite. It is used for the uniqueness.
A: There are many square roots for matrices, but for a real symmetric (or complex Hermitian) positive definite matrix, there's only one real symmetric and positive definite square root.
For example, $2\times 2$ square roots of the identity include every reflection matrix of the form $\begin{bmatrix}a&b\\b&-a\end{bmatrix}$ with $a^2+b^2=1$, but none of those are positive definite.
It's not quite that bad for most matrices; a $n\times n$ matrix with $n$ distinct positive eigenvalues only has $2^n$ square roots.
