# diagonalizability of matrix

If $A$ is invertible, $F$ is algebraically closed, and $A^n$ is diagonalizable for some $n$ that is not an integer multiple of the characteristic of $F$, then $A$ is diagonalizable.

My question is:

(1) Why the condition "$A$ is invertible" essential?

(2) In wikipedia, diagonalizable matrix , it says $A$ satisfies the polynomial $p(x)=(x^n- \lambda_1) \cdots (x^n- \lambda_k)$, is that means $p(x)$ is the minimal polynomail of $A^n$? If yes, why it can be written in the form of $(x^n- \lambda_1) \cdots (x^n- \lambda_k)$ instead of $(x- \lambda_1) \cdots (x- \lambda_k)$??

Thank you very much!

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A quick thought for (1): if $A$ is not invertible, then it might happen that $A$ is nilpotent, and then $A^n=0$ for some $n$, which is diagonalizable. But there are examples of nilpotent matrices which are not diagonalizable. Examples can already be found in the $2 \times 2$ case. – Nils Matthes Mar 3 '13 at 9:17

The polynomial you mention is not the minimal polynomial of $A^n$ itself, but rather closely related to it. Since $A^n$ is diagonalizable, its minimal polynomial splits into linear factors. Say $$p(x) = (x-\lambda_1)\cdots(x-\lambda_k)$$ where $p(A^n) = 0$. Of course this means that $A$ will satisfy $$q(x) = (x^n - \lambda_1)\cdots(x^n-\lambda_k)$$ since the two polynomials are related by $p(x^n)=q(x)$.

The condition that $A$ is invertible is essentially covered by Nils Matthes in the comments. The key point of the proof is that the above polynomial splits into distinct linear factors, forcing the minimal polynomial of $A$ to do the same.

By assumption of algebraic closure, the terms of the form $x^n - \lambda$ splits into distinct linear factors for $\lambda \neq 0$. But if $\lambda = 0$ then we simply have $x^n$ which does not split into distinct factors. This no longer forces the minimal polynomial to have distinct factors, in fact the minimal polynomial may contain a factor of $x$, or $x^2$, or $x^3$ or ... (you see where I'm going with this). If the minimal polynomial contains $x^k$ for $k>1$ then $A$ is not diagonalizable. Forcing $A$ to be invertible gets rid of this problem term.

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First note that a matrix is diagonalizable over the field $F$ if and only if its minimal polynomial is a product of distinct linear factors over $F$. Also a polynomial which is a product of distinct linear factors over $F$ is irreducible, and the polynomial $(x^n-\lambda)$ is a product of distinct linear factors over $F \iff \lambda\neq 0$ (use here that $n\nmid\text{char}(F)$).

Let's denote the matrix $A^n$ as $B=A^n$. If $B$ is diagonalizable then its minimal polynomial is $\mathcal m_x(B)=(x-\lambda_1)\cdot(x-\lambda_2)\cdots(x-\lambda_k)$ for some $\lambda_i\in F$ with $\lambda_i\neq\lambda_j$ for $i\neq j$. Therefore $$(A^n-\lambda_1)\cdot(A^n-\lambda_2)\cdots(A^n-\lambda_k)=(B-\lambda_1)\cdot(B-\lambda_2)\cdots(B-\lambda_k)=0.$$ Now use that $A$ is invertible to conclude that ($\lambda_i\neq 0$ and $(x^n-\lambda_1)\cdot(x^n-\lambda_2)\cdots(x^n-\lambda_k)$ is irreducible, so the polynomial ) $(x^n-\lambda_1)\cdot(x^n-\lambda_2)\cdots(x^n-\lambda_k)$ is the minimal polynomial of $A$. It follows (why?) that $A$ is diagonalizable.

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If $A$ is invertible then $A^n$ is also invertible, so $0$ is not an eigenvalue of $A^n$.

$A^n$ is diagonalizable then its minimal polynomial $P$ is a product of distinct linear factors over $F$: $$P(X)=(X-\lambda_1)\cdots(X-\lambda_k),\quad \lambda_i\neq\lambda_j$$ We know that $A^n$ is annihilated by $P$: $P(A^n)=0$ so $A$ is annihilated by the polynomial: $$Q(X)=(X^n-\lambda_1)\cdots(X^n-\lambda_k)$$ and since $\lambda_i\neq0$ then each polynomial $(X^n-\lambda_i)$ has no multiple root (since its derivative is $nX^{n-1}$ and $n$ is not an integer multiple of the characteristic of $F$), then $Q$ is a product of distinct linear factors and $A$ is annihilated by $Q$. Hence $A$ is diagonalizable matrix.

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