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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 \begin{equation} \displaystyle (u-\lambda_1\operatorname{Id})(u-\lambda_2\operatorname{Id})...(u-\lambda_k\operatorname{Id})=0. \end{equation}

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.

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  • $\begingroup$ When you say "free" does that mean "linearly independent"? $\endgroup$ – David C. Ullrich May 6 at 19:16
  • $\begingroup$ David Ullrich: yes. I don’t know if « free » is the exact traduction. (My course is in French) $\endgroup$ – 3809525720 May 6 at 19:26
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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 $k<n$, 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$$\alpha_0\operatorname{Id}+\alpha_1D+\cdots+\alpha_{n-1}D^{n-1}=0.$$

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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 $k<n$, this contradicts the linear independence of $\{\mathit{Id},u,\dots,u^{n-1}\}$.

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