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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.

Thanks,

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

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  • $\begingroup$ 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
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    $\begingroup$ 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
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Hints:

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

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