$A\in M_n(\mathbb C)$ 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?



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 $\lambda_i=\mu_i^m=\mu_j^m=\lambda_j$).

  • $\begingroup$ I don't understand. A is cancelled by which minimal polynomial? of $A^m$? $\endgroup$
    – Jozef
    Aug 22 '11 at 16:02
  • $\begingroup$ Sorry, I made a misprint. Now it's corrected. $\endgroup$
    – Henri
    Aug 22 '11 at 16:04
  • 1
    $\begingroup$ @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 $p(x)$. $\endgroup$ Aug 22 '11 at 16:06
  • 1
    $\begingroup$ @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). $\endgroup$ Aug 22 '11 at 16:37
  • 4
    $\begingroup$ @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. $\endgroup$ 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.


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