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Let $u \in \mathcal{L}(\mathbb{R}^5)$, with $u^3-u^2-u-2 \mathrm{Id}=0$ and $\mathrm{Tr}(u)=0$. So I have shown that $2$ is the only eigenvalue of $u$ and moreover that $\mathrm{Ker}(u^2+u+\mathrm{Id})\bigoplus\mathrm{Ker}(u-2\mathrm{Id})=\mathbb{R}^5$ with $\dim(\mathrm{Ker}(u-2\mathrm{Id}))=1$ therefore,

$$ \dim(\mathrm{Ker}(u^2+u+\mathrm{Id}))=4$$

How can I prove there exists a basis $(e_1,e_2,e_3,e_4)$ of $\mathrm{Ker}(u^2+u+\mathrm{Id})$ such that:

$$ \left\{\begin{array}{ll} u(e_1)=e_2\\ u(e_2)=-e_1-e_2\\ u(e_3) = e_4\\ u(e_4) = -e_3-e_4\end{array} \right.$$


My idea was:

Let $e_1 \neq0$ be an element of $\mathrm{Ker}(u^2+u+\mathrm{Id})$, then let's call $e_2 = u(e_1) $. I wanted to show afterwards that $(e_1,e_2)$ was linearly independant in $\mathrm{Ker}(u^2+u+\mathrm{Id})$, so I could take $e_3$ such that $(e_1,e_2,e_3)$ is still linearly independant in $\mathrm{Ker}(u^2+u+\mathrm{Id})$ and finally take $e_4 = u(e_3)$ so $(e_1,e_2,e_3,e_4)$ would verify all the conditions, except maybe the fact it is a basis of $\mathrm{Ker}(u^2+u+\mathrm{Id})$.


I don't know if it will work, and I feel there is something maybe more efficient, could somebody help me please?


In other words, my problem is as follows.

Let $A\in \mathcal{M}_5(\mathbb{R})$ be a matrix such that $\mathrm{Tr(A)=0}$ and $A^3-A^2-A-2I_n = 0$. I want to show that $A$ is similar to : \begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & -1 \end{pmatrix}

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I think that your approach is a good one.

Begin by taking $e_1$ to be any non-zero element of $\ker(u^2 + u + \operatorname{Id})$. To see that $e_2 = u(e_1)$ is linearly independent from $e_1$, suppose to the contrary that it is not. It follows that there exists a scalar $\lambda \in \Bbb R$ such that $u(e_1) = \lambda e_1$. Thus, $$ (u^2 + u + \operatorname{Id})e_1 = 0 \implies (\lambda^2 + \lambda + 1)e_1 = 0. $$ However, there is no real solution to the equation $\lambda^2 + \lambda + 1 = 0$, so this is impossible.

From there, we automatically satisfy part of the requirement for the basis. Indeed, we have $$ u(e_2) = u^2(e_1) = (-u - \operatorname{Id})e_1 = -e_1-e_2. $$ From there, take $e_3$ to be any element of $\ker(u^2 + u + \operatorname{Id})$ outside of the span of $e_1,e_2$, and let $e_4 = u(e_3)$. As before, we find that $e_3,e_4$ satisfy the required relations $$ u(e_3) = e_4, \quad u(e_4) = -e_3-e_4. $$ From there, one would need to show that $e_1,e_2,e_3,e_4$ is a linearly independent set, which is a bit tricky.


Suppose to the contrary that $e_4$ lies in the span of $e_1,e_2,e_3$. That is, there exist constants $c_1,c_2,c_3$ such that $$ e_4 = u(e_3) = c_1 e_1 + c_2 e_2 + c_3 e_3. $$ With that, we have $$ (u - c_3\operatorname{Id}) e_3 = c_1 e_1 + c_2 e_2. $$ We know that the polynomial $x^2 + x + 1$ has no real roots. Thus, the polynomials $x^2 + x + 1$ and $x - c_3$ are relatively prime. By Euclidean division, we have $$ x^2 + x + 1 = q(x)(x-c_3) + r $$ for some polynomial $q$ and some non-zero $r \in \Bbb R$. Equivalently, we have $$ q(x)(x - c_3) = r - (x^2 + x + 1) \implies\\ q(u)(x - c_3 \operatorname{Id})e_3 = [r \operatorname{Id} - (u^2 + u + \operatorname{Id})]e_3 = re_3. $$ So, we have $$ (u - c_3\operatorname{Id}) e_3 = c_1 e_1 + c_2 e_2 \implies\\ \frac 1r q(u)(u - c_3\operatorname{Id}) e_3 = \frac 1r q(u) (c_1 e_1 + c_2 e_2) \implies\\ e_3 = \frac 1r q(u) (c_1 e_1 + c_2 e_2). $$ However, $\frac 1r q(u) (c_1 e_1 + c_2 e_2)$ must be in the span of $e_1$ and $e_2$, so we have concluded that $e_3$ lies in the span of $e_1$ and $e_2$, which is false.


On the other hand, if you are already aware of the existence of rational canonical form, then you can reach the desired conclusion almost immediately.

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  • $\begingroup$ Thank you a lot. I had indeed trouble to prove $e_4 \not \in \mathrm{Span}(e_1,e_2,e_3)$. I am not aware of this rational canonical form but thanks anyway. $\endgroup$
    – Axel
    May 20 at 15:21
  • $\begingroup$ Alternatively, you can show that the cyclic subspace generated by $e_1$ has dimension at most two (and hence exactly two). $\endgroup$
    – lc2r43
    May 20 at 15:22
  • $\begingroup$ @lc2r43 Never heard of such concepts. But I'll take a look. $\endgroup$
    – Axel
    May 20 at 15:25
  • $\begingroup$ @lc2r43 How does this show that $e_4$ is not in the span of $e_1,e_2,e_3$? $\endgroup$ May 20 at 15:25
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    $\begingroup$ @Axel I've added a proof that $e_4 \notin \operatorname{Span}(e_1,e_2,e_3)$; see my latest edit. $\endgroup$ May 20 at 15:36
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This is my own solution to the following problem:

Let $A\in \mathcal{M}_n(\mathbb{R})$ be a matrix such that $\mathrm{Tr(A)=0}$ and $A^3-A^2-A-2I_n = 0$. I want to show that there exists $p \in \mathbb{N}$ such that $n=5p$ and that $A$ is similar to : $$C=\begin{pmatrix} 2I_p & 0 & 0 & 0 \\ 0 & B & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & B \end{pmatrix}$$ where $B = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}$

First of all, $P(X) = X^3-X^2-X-2$ annulates $A$ and one can see that:

$$P(X) = (X-2)(X^2+X+1) = (X-2)(X-j)(X-\bar{j})$$ where $j = \mathrm{e}^{i2 \pi/3}$.

Therefore as $P$ annulates $A$ and is irreducible with all its factors of degree $1$ and all of its roots with multiplicity equals to one thus I know $A$ is diagonalizable on $\mathbb{C}$.

Moreover, we know that $\sigma(A) \subset\{2,j,\bar{j}\}$ as $P$ annulates $A$.

Therefore, there exists $a,b,c \in\mathbb{N}$ such that its characteristic polynomial $\chi_A$ can be written:

$$\chi_A(X) = (X-2)^a(X-j)^b(X-\overline{j})^c$$

But $\chi_A \in \mathbb{R}[X]$ because $A$ is a matrix with real coefficients. Consequently $\chi_A = \overline{\chi_A}$ so $b = c$.

Moreover $\mathrm{Tr}(A) = 2a+bj+c\bar{j} = 2a+b(j+\overline{j})=2a-b=0 $, so $2a = b$. Finally $a+b+c = a+2b= \deg(\chi_A) = n$, we deduce that $\boxed{5a =n}$.

Therefore,

$$\chi_A(X) = (X-2)^{a}(X-j)^{2a}(X-\bar{j})^{2a}$$

As $a \neq 0$ we have $\sigma(A) = \{2,j,\bar{j}\} $ and because $A$ is diagonalizable on $\mathbb{C}$ there exists $P \in GL_n(\mathbb{C})$ such that,

$$A = P\begin{pmatrix} 2I_a & 0 & 0 & 0 & 0 & 0 \\ 0 & j & 0 & 0 & 0 & 0 \\ 0 & 0 & \bar{j} & 0 & 0 & 0 \\ 0 & 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & 0 & 0 & j & 0 \\ 0 & 0 & 0 & 0 & 0 & \bar{j} \end{pmatrix} P^{-1}$$

On top of that if we call $B = \begin{pmatrix} 0 & -1 \\ 1 & -1 \end{pmatrix}$ we have $\chi_B(X)= X^2+X+1 = (X-j)(X-\bar{j})$, thus because $j \neq \overline{j}$ we can find $Q \in GL_2(\mathbb{C})$ such that,

$$ B = Q \begin{pmatrix} j & 0 \\ 0 & \overline{j} \end{pmatrix} Q^{-1} $$

Hence,

$$C = \begin{pmatrix} I_a & 0 & 0 & 0 \\ 0 & Q & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & Q \end{pmatrix} \begin{pmatrix} 2I_a & 0 & 0 & 0 & 0 & 0 \\ 0 & j & 0 & 0 & 0 & 0 \\ 0 & 0 & \bar{j} & 0 & 0 & 0 \\ 0 & 0 & 0 & \ddots & 0 & 0 \\ 0 & 0 & 0 & 0 & j & 0 \\ 0 & 0 & 0 & 0 & 0 & \bar{j} \end{pmatrix} \begin{pmatrix} I_a & 0 & 0 & 0 \\ 0 & Q & 0 & 0 \\ 0 & 0 & \ddots & 0 \\ 0 & 0 & 0 & Q \end{pmatrix}^{-1}$$

Consequently, $A$ is similar to the real matrix $C$ in $\mathcal{M}_n(\mathbb{C})$ so we can write $A = RCR^{-1}$ with $R = R_1 +iR_2\in GL_n(\mathbb{C})$ and $R_1,R_2 \in \mathcal{M}_n(\mathbb{R})$

Hence $AR = RC$, then $ AR_1 = R_1C$ and $AR_2 = CR_2$.

Moreover $\varphi : x \mapsto \det(R_1+xR_2) $ is a polynomial function with real coefficients and we know $\varphi(i) = \det(R) \neq 0$ therefore $\varphi$ has a finite number of roots, meaning we can find $x_0 \in \mathbb{R}$ such that $\varphi(x_0) \neq0$, id est $R_0 = R_1+x_0R_2 \in GL_2(\mathbb{R})$ and :

$$AR_0 = AR_1+x_0AR_2 = R_1C +x_0R_2C = R_0C$$

Finally,

$$A = R_0 C R_0^{-1}$$

So it proves that $A$ is similar to $C$ in $\mathcal{M}_n(\mathbb{R})$.

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  • $\begingroup$ Well done! The argument via the determinant is very nice. Thinking about it, I remembered that I used a similar argument on my post here. You might also find this post about similarity and field extensions (which uses a generalized version of your argument) to be interesting $\endgroup$ May 20 at 22:28
  • $\begingroup$ @BenGrossmann Thank you for your time, and for the interesting links. I still have a lot to learn! $\endgroup$
    – Axel
    May 21 at 4:12
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Let us think of the problem in terms of the associated $\mathbb{R}[u]$-module. The condition that $u^3 - u^2 - u - 2 = (u-2) (u^2+u+1)$ annihilates this module implies that the decomposition of the module into a direct sum of indecomposable cyclic modules must have factors $\mathbb{R}[u] / \langle u-2 \rangle$ and $\mathbb{R}[u] / \langle u^2+u+1 \rangle$ only. Then, the restriction that the dimension as a vector space over $\mathbb{R}$ is 5 restricts us to the choices $(\mathbb{R}[u] / \langle u-2 \rangle)^5$, $(\mathbb{R}[u] / \langle u-2 \rangle)^3 \oplus \mathbb{R}[u] / \langle u^2+u+1 \rangle$, or $\mathbb{R}[u] / \langle u-2 \rangle \oplus (\mathbb{R}[u] / \langle u^2+u+1 \rangle)^2$; and out of these, the only one with $\operatorname{tr}(u) = 0$ is the last one. Now, $u$ operating on this module $\mathbb{R}[u] / \langle u-2 \rangle \oplus (\mathbb{R}[u] / \langle u^2+u+1 \rangle)^2$ has $\ker(u^2+u+\operatorname{id}) = 0 \oplus (\mathbb{R}[u] / \langle u^2+u+1 \rangle)^2$, and $e_1 := (0, 1, 0), e_2 := (0, u, 0), e_3 := (0, 0, 1), e_4 := (0, 0, u)$ does form an $\mathbb{R}$-basis of $\ker(u^2+u+\operatorname{id})$ with the desired properties.

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