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I'm asked to prove that the matrix $A\in M_{n}(\mathbb C)$ that satisfy $A^8+A^2=I$ is diagonalizable. I've tried looking at the equation $x^8+x^2-1=0$ and determining whether $M_A$ has any repeating roots, but this got me nowhere. Afterwards, I thought about trying to determine whether its Jordan form is diagonal (I know such form exists since $\mathbb C$ is algebraically closed, so $P_A$ splits into linear factors) still got nowhere. Is there a right approach to the question or is there something I'm missing?

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Is it not enough to check if this polynomial has a double root?

Its derivative is $$ 8x^7 + 2x = x(8x^6 + 2) = 8x(x^6 + \frac 14). $$ Now $0$ is not a common root, so the double roots would satisfy $$ x^6 + \frac14 = 0. $$ Plugging this into our original polynomial, a double root would satisfy $$ 0 = x^8 + x^2 - 1 = x^2(x^6 + \frac14) + \frac34 x^2 -1= \frac34 x^2 -1, $$ but the roots of $x^2 - \frac43$ are not roots of $x^6 + \frac14$, since if $x^2 = \frac 43$, then $$ x^6 + \frac14 = \left(\frac43\right)^3 + \frac14 \neq 0. $$

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    $\begingroup$ I do deem this as enough. Thank you very much. $\endgroup$
    – GBA
    May 24 '20 at 20:03
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I think your first idea to check the roots of $f(x) = x^8 + x^2 - 1$ is good. With some careful arguments you can show that $f$ has distinct roots.

Let $\alpha$ be a root and let $\beta = \alpha^2$. So we have $\beta^4 + \beta - 1 = 0$. The polynomial $x^4 + x - 1$ has two distinct real roots $a < 0 < b$ and a pair of complex conjugate roots $c,d$. So there are four possible choices for $\beta$. Since $b > 0$ we have that $\sqrt{b}$ and $-\sqrt{b}$ are real roots of $f$. Since $a < 0$ we have that $i\sqrt{-a}$ and $-i\sqrt{-a}$ are also distinct solutions for $f$

When we take the squareroot of $c$ and $d$, we need to be sure that none of these numbers coincides with any of the roots of $f$ we've found so far. Since $c$ and $d$ are complex conjugates, let's write them as $c = re^{i\theta}$ and $d = re^{-i \theta}$ for some $r > 0$ and $\theta \in (0, \pi)$. Note that this choice of $\theta$ is possible because $c, d$ are not real. Taking their squareroots we get the following roots of $f$: $$ \sqrt r e^{i (\theta/2)}, \sqrt re^{-i (\theta/2)}, \sqrt r e^{i (\pi/2 + \theta/2)}, \sqrt r e^{-i (\pi/2 + \theta/2)}. $$ If you plot these out in the complex plane you will notice they all lie in different quadrants and none of them are purely real or imaginary. So all the roots are distinct, this shows that the minimal polynomial has distinct roots and so $A$ is diagonalisable.

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    $\begingroup$ The roots of your $f$ are not eigen-values. $\endgroup$ May 23 '20 at 22:20
  • $\begingroup$ Thaks for the comment, for some reason I had thought $f$ was the characteristic polynomial $\endgroup$ May 23 '20 at 22:35
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Look at $g(x)=x^4+x-1$ . Observe that $g'(x)=4x^3+1$ has only one real root and deduce that g as no repeated root in $\mathbb R$.

Moreover, we have $g(0)=-1$ and $g(x)\to \infty$ for $x\to \infty$ and $x\to -\infty$.

So $g$ has only two distinct real roots. The other two roots are distinct and complex occurring in conjugate pairs.

Now we have $f(x)=x^8+x^2-1=g(x^2)$

$\therefore \{\alpha \in \mathbb C|f(x)=0\}=\{\alpha \in \mathbb C|\alpha^2 \text{ is a root of }g\} $

As the roots of $g$ are distict so are the roots of $f$. So $f$ is product of distinct linear factors.

Now you know your minimal polynomial $M_A$ divides $f$ in $\mathbb C[x]$ as $f$ annihilates $A$.

Hence $M_A$ is just the product of distinct linear factors hence $A$ is diagonalizible.

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