Let $A\in M_{n,n}(\mathbb{C})$ be a diagonalisable matrix. Prove $\exists B \in M_{n,n}(\mathbb{C})$ such that $B^{2016} = A$

Question

Let $A\in M_{n,n}(\mathbb{C})$ be a diagonalisable matrix.

Prove $\exists B \in M_{n,n}(\mathbb{C})$ such that $B^{2016} = A$

I can't see why this statement would be true. Perhaps I'm missing something.

Can you assist me please?

Answer 1

Hint: Since $A$ is diagonalizable, there exists an invertible matrix $P$ such that $P^{-1}AP = \text{diag}(\lambda_{1}, ..., \lambda_{n})$.

$\lambda_{j} = r_{j}e^{i\theta_{j}}$ and there are $n$ distinct numbers: $u_{k} = r_{j}^{\frac{1}{n}}e^{i\left(\frac{\theta_{j} + 2 \pi k}{n}\right)}$