Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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

3 Answers 3

up vote 1 down vote accepted

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.

share|improve this answer

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.

share|improve this answer
    
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 $e_i$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.