# 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.

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Well, why not? For different decompositions $A=U_1 \Lambda_1 U_1 ^t = U_2 \Lambda_2 U_2 ^t$, it might be possible that $U_1 \Lambda_1^{1/k} U_1^t=U_2 \Lambda_2^{1/k} U_2 ^t$, and in fact this turns out to be the case (assuming $U_1,U_2$ to be unitary matrices of course). –  Florian Apr 4 '11 at 15:07
Five seconds difference! :) –  Glen Wheeler Apr 4 '11 at 15:07
In my version of Horn, uniqueness is proved in the following paragraph. Sure, the decomposition $A=U \Lambda U^T$ is not unique, but just because changing this decomposition will change the given decomposition of $B$ does not mean that $B$ itself will be changed. –  Barry Smith Apr 4 '11 at 15:07
Thanks everyone. I should notice that $U$ is different but $U\Lambda^{1/k}U^T$ is invariant. –  Shiyu Apr 4 '11 at 15:12
It should be noted that there are at least $k^n$ $k$'th roots (over the complex numbers) of an $n \times n$ symmetric matrix A (and may be infinitely many), but only one of them can be positive definite. –  Robert Israel Apr 4 '11 at 15:26

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.

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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.

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Is "$U\Lambda U^T$ is also unique" a convoluted way of saying "A is the unique matrix with the property of being equal to A"? :) –  wildildildlife Apr 4 '11 at 20:56
Yes, it is ;). I guess I could have put more effort into this answer. –  Glen Wheeler Apr 4 '11 at 21:10
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 $e_i$.