# Why this endomorphism has n distinct eigenvalues?

Let $$u$$ be a diagonalisable endomorphism of $$\mathbb{R}^n$$. Let’s suppose that the family $$(\operatorname{Id}, u, u^2,..., u^{n-1})$$ is linearly independent and consider $$\lambda_1, \lambda_2,..., \lambda_k$$, the pairwise distinct eigenvalues of $$u$$.

Because $$u$$ is diagonalisable $$$$\displaystyle (u-\lambda_1\operatorname{Id})(u-\lambda_2\operatorname{Id})...(u-\lambda_k\operatorname{Id})=0.$$$$

This is a polynomial in $$u$$ whose degree is $$k$$. In order to show that $$k=n$$, I have to justify that $$k > n-1$$ using the above relation.

I have difficulties to visualize precisely the situation of a nontrivial linear dependency between $$\operatorname{Id}, u, u^2,...,u^{n-1}$$. Many thanks for any help.

• When you say "free" does that mean "linearly independent"? – David C. Ullrich May 6 at 19:16
• David Ullrich: yes. I don’t know if « free » is the exact traduction. (My course is in French) – 3809525720 May 6 at 19:26

Let $$B=\{v_1,v_2,\ldots,v_n\}$$ a basis of $$\Bbb R^n$$ such that the matrix of $$u$$ with respect to $$B$$ is a diagonal matrix $$D$$. Then each entry of the main diagonal of $$D$$ is some $$\lambda_j$$. If there are less than $$n$$ of them, then there will be repeated entries on the main diagonal. But then $$\{\operatorname{Id},D,D^2,\ldots,D^{n-1}\}$$ cannot be linearly independent. In fact$$(1,1,\ldots,1),(\lambda_1,\lambda_2,\ldots,\lambda_k),\ldots,\left(\lambda_1^{\,n-1},\lambda_2^{\,n-1},\ldots,\lambda_k^{\,n-1}\right)$$are $$n$$ vectors of $$\Bbb R^k$$ and therefore, if $$k, they are linearly dependent, that is, there are scalars $$\alpha_0,\ldots,\alpha_{n-1}$$, not all of which are $$0$$, such that$$\alpha_0(1,1,\ldots,1)+\alpha_1(\lambda_1,\lambda_2,\ldots,\lambda_k)+\cdots+\alpha_{n-1}\left(\lambda_1^{\,n-1},\lambda_2^{\,n-1},\ldots,\lambda_k^{\,n-1}\right)=0.$$In other words, if $$j\in\{1,2,\ldots,k\}$$, then$$\alpha_0+\alpha_1\lambda_j+\cdots+\alpha_{n-1}\lambda_j^{\,n-1},$$and therefore$$\alpha_0\operatorname{Id}+\alpha_1D+\cdots+\alpha_{n-1}D^{n-1}=0.$$
The polynomial $$f(x)=(x-\lambda_1)(x-\lambda_2)\dotsm(x-\lambda_k)$$ is monic, hence nonzero, and $$f(u)=0$$. This yields a linear dependence relation between $$\mathit{Id},u,\dots,u^k$$. If $$k, this contradicts the linear independence of $$\{\mathit{Id},u,\dots,u^{n-1}\}$$.