Prove that $B^n$ is diagonalisable for all $n=2,3,\dots$ and that every eigenvalue of $B^2$ is the square of some eigenvalue of $B$.

Question

I would like to ask you for some help in the following problem:

Suppose that a matrix $B$ is diagonalisable over $\mathbb{C}$. Prove that $B^n$ is diagonalisable for all $n=2,3,\dots$ and that every eigenvalue of $B^2$ is the square of some eigenvalue of $B$.

Has anyone got some suggestions?

Answer 1

If $B$ is diagonalizable then $\exists \, P$ (an invertible matrix and $D$ a diagonal matrix such that $B=PDP^{-1}$. Then $B^n=(PDP^{-1})^n=PD^nP^{-1}$. Thus $B^n$ is similar to a diagonal matrix $D^n$, hence diagonalizable.

Let $\lambda$ be an eigen value of $B$, this means $\exists$ a non-zero vector $\mathbf{v}$ such that $B\mathbf{v}=\lambda\mathbf{v}$. Now $$B^2\mathbf{v}= B(B\mathbf{v})=B(\lambda\mathbf{v})=\lambda B\mathbf{v}=\lambda^2 \mathbf{v}.$$