Let A be nxn with real coefficients and assume that it has n distinct eigenvalues, and all eigenvalues are positive real numbers.

Let k $\ge$3 be an odd integer.

a) Prove there exists a unique real matrix B with $B^k$ = A.

b) How many complex matrices B satisfy $B^k$ = A. (Include the real matrices B in your count.)

I would like hints only.

I know, from my previous questions on MSE, that positive real eigenvalues does not imply A is symmetric, and hence not necessarily positive-definite.

I also know that A is diagonalizable (not necessarily orthogonally diagonalizable), because of the n distinct eigenvalues it has - so $$A = SDS^{-1}$$,

where S's columns are eigenvectors of A.

...now I need to somehow make use of the other information given, namely that the eigenvalues are positive and real.



Hint: $B$ is diagonalized by the same $S$.

  • $\begingroup$ Thanks, Professor Israel. It is $SD^\frac{1}{k}S^{-1}$, and to check uniqueness, we need only check that D^(1/k) is unique, but the diagonal entries are unique for k odd, since we look at the relation between B and A's eigenvalues and solve $x_{ii}^\frac{1}{k} = d_{ii}$. Can you offer a hint for part (b)? Seems strange that after we've established that there is a unique B that now we are asked to count many more (complex) matrices that satisfy $B^k = A$. Thanks, $\endgroup$ – User001 Jul 3 '15 at 3:20
  • 1
    $\begingroup$ Instead of the positive $k$'th root, use ... $\endgroup$ – Robert Israel Jul 3 '15 at 3:36
  • $\begingroup$ complex k'th roots :-). Thanks so much, Professor Israel - have a great night! $\endgroup$ – User001 Jul 3 '15 at 5:14


  • $B$ is diagonalized by the same $S$.
  • The entries of $D$ correspond to the eigenvalues of $A$.
  • $\begingroup$ Thanks, @bashfuloctopus. Can you offer a hint for part(b)? :-) $\endgroup$ – User001 Jul 3 '15 at 3:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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