TITLE: Square root of Positive Definite Matrix
QUESTION [17 upvotes]: Let $A$ be an $n\times n$ positive definite matrix. Show that there exists a unique positive definite matrix $B$ such that $B^2=A$.
I do know the existence. But what about the uniqueness? Would you help me out? Thank you.

REPLY [14 votes]: Being symmetric, $A$ is diagonalisable. Any square root $B$ commutes with $A$, so it must stabilise the eigenspaces of $A$. The restriction of $B$ to the eigenspace $V$ of $A$ for $\lambda>0$ is a symmetric positive definite square root of the restriction $\lambda I_V$ of $A$, so it suffices to show that $\sqrt\lambda I_V$ is the unique such square root.
But the restriction of $B$ is diagonalisable, any eigenvalue it has must be a square root of$~\lambda$ and also positive; a diagonalisable matrix with $\sqrt\lambda$ as unique eigenvalue cannot be other than $\sqrt\lambda I_V$.