Uniqueness of the kth root of a positive definite matrix Let $A$ be a positive definite matrix. $B$ is said to be the $k$th root of $A$ if $B^k=A$.
My question is whether $B$ is unique.
In Matrix Analysis, Horn, 7.2.6, it is stated 'there exists a unique positive definite Hermitian matrix B such that $B^k=A$'. It seems that $B$ can be $B=U\Lambda^{1/k}U^T$ if $A=U\Lambda U^T$. But the decomposition $A=U\Lambda U^T$ is not unique in general, right? Hence $B=U\Lambda^{1/k}U^T$ also should not be unique.
 A: Although the decomposition is not unique, the diagonal form of $A$ (that is $\Lambda$) is unique up to the order of the diagonal entries. Thus, if 
$$ A = Q\Psi Q^T$$
is another decomposition of $A$ with $\Psi$ diagonal and $Q$ orthogonal, then $\Psi$ and $\Lambda$ are related by permuting rows and columns; that is, there is a matrix $R$ which is obtained by permuting the columns of the identity matrix (and in particular, $R$ is orthogonal) such that $R\Psi R^{-1} = R\Psi R^T = \Lambda$.
Note that $A = U\Lambda U^T = U(R\Psi R^T)U^T = (UR)\Psi(UR)^T = Q\Psi Q^T$.
It is then straightforward to check that $R\Psi^{1/k}R^{T} = \Lambda^{1/k}$, and that $(UR)\Psi^{1/k}(UR)^T = Q\Psi^{1/k}Q^T$. 
So if you pick the decomposition $Q\Psi Q^T$ instead of $U\Lambda U^T$, then the positive definite $k$th root of $A$ you get will be
$$B = Q\Psi^{1/k}Q^{-1}.$$
But 
$$U\Lambda^{1/k}U^T = U(R\Psi^{1/k}R^T)U^T = (UR)\Psi^{1/k}(UR)^T = Q\Psi^{1/k}Q^T;$$
that is, the matrix you get is actually equal to the one you got originally.
A: Matrix roots are in general not unique (or even defined), but in this case they are.
In the decomposition $A = U\Lambda U^T$, $U$ is not unique, but $\Lambda = \text{diag}(\lambda_1,\ldots,\lambda_n)$ is unique.  (The product $U\Lambda U^T$ is also unique.)  Thus, and more to the point, the roots $\Lambda^{1/k} = \text{diag}(\lambda_1^{1/k},\ldots,\lambda_n^{1/k})$ are unique.  This means $B$ is unique.
A: The decomposition $A = U\Lambda U^T$ is indeed not unique but it only depends on the order of a basis $\mathcal B = \{e_1, \dots e_n\}$ of eigenvectors of $A$. If you look closely at how $B$ is defined, you will notice that $B$ corresponds to the unique linear transformation mapping $e_i \mapsto \lambda_i^{1/k}e_i$ for all $i \in \{1,\dots,n\}$, where $\lambda_i$ is the eigenvalue of $A$ corresponding to $e_i$.
