# Prove there exists a particular basis of $\mathrm{Ker}(u^2+u+Id)$

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 : $$\begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 0 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 & 0 \\ 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 1 & -1 \end{pmatrix}$$

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.

• 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.
– Axel
May 20 at 15:21
• Alternatively, you can show that the cyclic subspace generated by $e_1$ has dimension at most two (and hence exactly two). May 20 at 15:22
• @lc2r43 Never heard of such concepts. But I'll take a look.
– Axel
May 20 at 15:25
• @lc2r43 How does this show that $e_4$ is not in the span of $e_1,e_2,e_3$? May 20 at 15:25
• @Axel I've added a proof that $e_4 \notin \operatorname{Span}(e_1,e_2,e_3)$; see my latest edit. May 20 at 15:36

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 $$\mathcal{M}_n(\mathbb{R})$$.

• 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 May 20 at 22:28
• @BenGrossmann Thank you for your time, and for the interesting links. I still have a lot to learn!
– Axel
May 21 at 4:12

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.