# Showing that there is unique matrix $B$ such that $B^k=A$ for some $A$

Let $$A$$ be a $$n$$ by $$n$$ real matrix with distinct positive eigenvalues $$\lambda_1$$,...,$$\lambda_n$$. And let $$k$$ be an odd integer. Then, I was able to show that there exists a real matrix $$B$$ such that $$B^k=A$$. However, it is not so easy to show that such $$B$$ is unique. How do I exclude the possibility that $$B$$ has some complex eigenvalues and still the entries of $$B$$ are all real?

$$B$$ is a real matrix. If it has some non-real eigenvalues, these eigenvalues must appear in conjugate pairs. In turn, $$B$$ has two eigenvalues of identical moduli and so does $$A=B^k$$. But this is impossible, because by assumption, $$A$$ has a real spectrum consisting of distinct positive numbers.
Thus, if $$B$$ is a real matrix $$k$$-th root of $$A$$, it must have a real spectrum. Since $$k$$ is odd, the eigenvalues of $$B$$, being the $$k$$-th roots of the eigenvalues of $$A$$, must be distinct positive real numbers.
Hence $$B$$ has a diagonalisation $$B=PDP^{-1}$$ over $$\mathbb R$$. Let $$f$$ be any polynomial such that $$f(\lambda_i)=\lambda_i^{1/k}$$ for each eigenvalue $$\lambda_i$$ of $$A$$. Then $$f(D^k)=D$$. In turn, $$f(A)=f(B^k)=f(PD^kP^{-1})=Pf(D^k)P^{-1}=PDP^{-1}=B.$$ Therefore $$B$$ is unique, because it is necessarily equal to $$f(A)$$ (and $$f$$ depends only on the multiset of eigenvalues of $$A$$ but not on $$P$$). This actually also proves the existence of $$B$$: you just pick $$B=f(A)$$.
• I can't finish. How do we know that there isn't some real matrix $P$ commuting with $A$ but not $B$? If there is, then $PBP^{-1}$ is another $k^{th}$ root of $A$, since $(PBP^{-1})^k = PAP^{-1} = A$. – mathworker21 Aug 17 at 22:14
• @mathworker21: Alternatively, since $B$ has distinct real eigenvalues, it is diagonalizable with respect to its eigenbasis, which we can directly confirm is also an eigenbasis for $A.$ With respect to this basis, both $A$ and $B$ are diagonal, each with distinct entries along their diagonals. What type of matrix can commute with a diagonal matrix with distinct entries? – Brian Moehring Aug 17 at 23:04