If given $A^n = \alpha I$ prove $A$ is diagonizable

Question

Given matrix $ A \in{M_n(\mathbb{C})}, 0\ne\alpha\in{\mathbb{C}}$ such that $A^n = \alpha I$ prove that $A$ is diagonizable.

I've tried proving that the minimal polynomial($m_A$) splits into linear polynomials therefore proving that $A$ is diagonizable but haven't had great success with that.

I've also found that $|A| = \alpha$ even though that didn't really give me a good starting place.

How can I prove that $A$ is diagonizable?

Answer 1

Small remark: Over $\mathbb{C}$, every polynomial splits into linear factors, so that doesn't help here. The matrix is diagonizable if and only if the minimal polynomial splits in pairwise different linear factors, so it splits and is square free.

Now for your problem: We see directly by looking at the degree that the characteristic polynomial is $x^n-\alpha$. Thus, the minimal polynomial is a divisor of that one. Show that this polynomial is square free, then also the minimal polynomial, a divisor of the characteristic polynomial, will be square free and you are done.

Answer 2

If $A^n=\alpha\operatorname{Id}$, with $\alpha\neq0$, then $0$ is not an eigenvalue of $A$. Suppose that $A$ is not diagonalizable. Then some block of the Jordan normal form of $A$ is of the type$$\begin{bmatrix}\lambda&1&0&0&\ldots&0\\0&\lambda&1&0&\ldots&0\\0&0&\lambda&1&\ldots&0\\\vdots&\vdots&\vdots&\vdots&\ddots&\vdots\\0&0&0&0&\cdots&\lambda\end{bmatrix},$$with $\lambda\neq0$. But no power of such a matrix is diagonlizable.

Answer 3

Hint: the essential fact is that eigenvectors corresponding to different eigenvalues are linearly independent. In this case, we have a basis of the space composed of eigenvectors...