# $A$ is diagonalizable, then exists $B^3=A$

I got this question on my linear algebra exam. The question was to prove/disprove the following statement:

If $A$ is a diagonalizable matrix. Then there exists a matrix $B$ so that $B^3 = A$.

I tried using the definition of diagonalizable matrices, but without any success.

• The field is relevant. Presumably the real numbers are intended. (It could be instructive to consider the same problem with $B^2 = A$) – quid Aug 8 '18 at 23:25

This is true. Diagonalize $A = P \Lambda P^{-1}$ and let $B = P\Gamma P^{-1}$ with $\Gamma$ being the diagonal matrix with $\Gamma_{ii} = \Lambda_{ii}^{1/3}$. Then $\Gamma^3=\Lambda$ and hence $$B^3 = (P\Gamma P^{-1})\,(P\Gamma P^{-1})\,(P\Gamma P^{-1}) = P\Gamma^3 P^{-1} = P\Lambda P^{-1} = A.$$