Square of two positive definite matrices are equal then they are equal I have read that if $P, Q$ are two positive definite matrices such that $P^2=Q^2$, then $P=Q$.
I don't know why. Some one can help me? Thanks for any indication.
 A: It all boils down to this:

Proposition. Suppose $A$ is an $n\times n$ positive definite matrix. Then $A^2$ has an eigenbasis. Furthermore, given any eigenbasis $\{v_1, \ldots, v_n\}$ of $A^2$ such that for each $i$, $A^2v_i=\lambda_iv_i$ for some $\lambda_i>0$, we must have $Av_i=\sqrt{\lambda_i}v_i$.

I will leave the proof of this proposition to you. Now, suppose $\{v_1,\ldots,v_n\}$ is an eigenbasis for $P^2=Q^2$ with $P^2v_i=Q^2v_i=\lambda_iv_i$. By the above proposition, we must also have $Pv_i=Qv_i=\sqrt{\lambda_i}v_i$. Since the mappings $x\mapsto Px$ and $x\mapsto Qx$ agree on a basis of the underlying vector space, $P$ must be equal to $Q$.
A: *

*If P and Q are symmetric positive definite, and $P^2=Q^2$, then the eigenvalues of the squared matrices and their eigenvectors are the same. 
(The characteristic function is the same)

*If the eigenvalues of $P^2$ are all positive, the eigenvalues of $P$ are merely the (real) squareroots of the eigenvalues of $P^2$.

*The eigenvectors of P and Q are the same as that of $P^2$ and $Q^2$. 

*By constructing P and Q using the $SAS^{-1}$ form, you can prove that they are equal.

*Note that symmetric matrices are never defective.

A: A positive semidefinite matrix has a unique positive semidefinite square root. 
http://www.youtube.com/watch?v=6hQqFYVGQfs
