# $A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. Prove $A^{2005}$ is not diagonalizable

$A\in M_n(\mathbb C)$ invertible and non-diagonalizable matrix. I need to prove that $A^{2005}$ is not diagonalizable as well. I am asked as well if Is it true also for $A\in M_n(\mathbb R)$. (clearly a question from 2005).

This is what I did: If $A\in M_n(\mathbb C)$ is invertible so $0$ is not an eigenvalue, We can look on its Jordan form, Since we under $\mathbb C$, and it is nilpotent for sure since $0$ is not an eigenvalue, and it has at least one 1 in it's semi-diagonal. Let $P$ be the matrix with Jordan base, so $P^{-1}AP=J$ and $P^{-1}A^{2005}P$ but it leads me nowhere.

I tried to suppose that $A^{2005}$ is diagonalizable and than we have this $P^{-1}A^{2005}P=D$ When D is diagonal and we can take 2005th root out of each eigenvalue, but how can I show that this is what A after being diagonalizable suppose to look like, for as contradiction?

Thanks

If $A^m$ is diagonalizable, then $A^m$ is cancelled by its minimal polynomial $P$, which has simple roots. Therefore $A$ is cancelled by $P(X^m)$ which has simple roots because $P(0)\neq 0$ ($A$ is invertible).

Indeed, if $P(X)=\prod (X-\lambda_i)$, then $P(X^m)=\prod (X^m-\lambda_i)$ whose roots are all the $m$-roots of $\lambda_i$ which differ one frome another (if $\mu_i$ is a $m$-root of $\lambda_i$, then $\mu_i\neq \mu_j$ else $\lambda_i=\mu_i^m=\mu_j^m=\lambda_j$).

• I don't understand. A is cancelled by which minimal polynomial? of $A^m$? Aug 22 '11 at 16:02
• Sorry, I made a misprint. Now it's corrected. Aug 22 '11 at 16:04
• @user14829: It means that if $\mu(x)$ is the minimal polynomial of $A^m$, then $\mu(A^m)$. Write out $\mu(x)=x^k + a_{k-1}x^{k-1}+\cdots+a_1x + a_0$. Plug in $A^m$. You get $A^{mk} + a_{k-1}A^{m(k-1)} + \cdots+a_1A^{m} + a_0I$. This is the same as you would get if you evaluate $p(x) = x^{mk} + a_{k-1}x^{m(k-1)} + \cdots + a_1x^m + a_0$ at $A$, so $p(A)=0$, so the minimal polynomial of $A$ must divide $p(x)$. Aug 22 '11 at 16:06
• @Jozef: $p(x)$ splits into distinct linear factors because the minimal polynomial of $A^m$ is a product of linear factors $(x-\lambda_i)$, where $\lambda_i\neq 0$ are the distinct eigenvalues; so $p(x)$ is a product of polynomials $(x^m-\lambda_i)$, which split over $\mathbb{C}$ with no repeated roots (because $\lambda_i\neq 0$), and no common factors (because the $\lambda_i$ are pairwise distinct). $p(x) = \mu(x^m)$. You use invertibility of $A$ when you know that no $\lambda_i$ is zero (if $\lambda_i=0$, then you get $x^m$, and this one gives you repeated roots). Aug 22 '11 at 16:37
• @Jozef: Over $\mathbb{R}$ the result is false. Take the rotation by $90^{\circ}$, $A=\left(\begin{array}{rr}0&1\\-1&0\end{array}\right)$. Then $A^2=-I$ is diagonalizable, but $A$ is not. Even with all eigenvalues positive for $A^m$ the result does not follow for $m\gt 2$, because the factors $x^m-\lambda$ no longer split, since the real numbers have at most two roots for $x^m-\alpha$. It's true if $A^2$ is diagonalizable and has all positive eigenvalues. Aug 22 '11 at 16:48

As luck would have it, the implication: $A^n$ diagonalizable and invertible $\Rightarrow A$ diagonalizable, was discussed in XKCD forum recently. See my answer there as well as further dicussion in another thread.

This is just a rewriting of this other answer adding some more details in the derivation.

Suppose $$A^m$$ is diagonalisable. This is equivalent to the minimal polynomial of $$A^m$$ splitting into linear factors. Denoting with $$\mu$$ the minimal polynomial of $$A^m$$ and with $$\lambda_i$$ its eigenvalues (where $$\lambda_i\neq\lambda_j$$ for $$i\neq j$$), this means that the polynomial $$\mu(x) = \prod_i (x - \lambda_i)\tag A$$ is the one with the smallest degree such that $$\mu(A^m)=0$$.

Define now the polynomial $$p$$ as the one such that $$p(x)=\mu(x^m)$$. Clearly, this satisfies $$p(A)=0$$. Using the expression (A), we can write $$p$$ as $$p(x)=\prod_i (x^m - \lambda_i).\tag B$$

Now, if we are working in $$\mathbb C$$, then all polynomials of the form $$x^m-c$$ with $$c\neq0$$ have $$m$$ distinct roots and therefore split linearly as $$x^m - c = \prod_j (x - c_j), \qquad c_j^m = c.$$

Note that if $$A$$ is non-invertible, this leaves open the possibility that $$\lambda_i=0$$ for some $$i$$, and then $$p$$ contains a factor of the form $$x^m$$, which would imply $$A$$ is not diagonalisable (it doesn't have to imply this, because the minimal polynomial of $$A$$ only has to divide $$p$$, and therefore could still not contain these terms).

If, on the other hand, $$A$$ is invertible, then we know that $$\lambda_i\neq0$$, therefore $$p$$ splits linearly, and therefore the minimal polynomial of $$A$$ does as well.