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?

Could anyone please explain?


1 Answer 1


$B$ 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 $B=f(A)$.

  • $\begingroup$ 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 $(PBP^{-1})^k = PAP^{-1} = A$. $\endgroup$ Commented Aug 17, 2019 at 22:14
  • $\begingroup$ @mathworker21 See my new edit. $\endgroup$
    – user1551
    Commented Aug 17, 2019 at 22:44
  • 1
    $\begingroup$ @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? $\endgroup$ Commented Aug 17, 2019 at 23:04
  • $\begingroup$ @BrianMoehring yes, I didn't write out to see what can commute with a diagonal matrix with distinct entries. thanks. $\endgroup$ Commented Aug 17, 2019 at 23:44

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