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Linear Projection

Linear Algebra
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Riemann
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Linear Projection

#1

Post by Riemann » Sun Sep 22, 2019 5:36 pm

Let $\mathcal{V}$ be a linear space over $\mathbb{R}$ such that $\dim_{\mathbb{R}} \mathcal{V} < \infty$ and $f:\mathcal{V} \rightarrow \mathcal{V}$ be a linear projection such that any non zero vector of $\mathcal{V}$ is an eigenvector of $f$. Prove that there exists $\lambda \in \mathbb{R}$ such that $f = \lambda \; \mathrm{Id}$ where $\mathrm{Id}$ is the identity endomorphism.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$

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S.F.Papadopoulos
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Re: Linear Projection

#2

Post by S.F.Papadopoulos » Sat Sep 28, 2019 6:34 pm

$f=I$
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Re: Linear Projection

#3

Post by Riemann » Mon Sep 30, 2019 2:54 pm

Hi ,

I'm sorry but I do not understand what exactly you wrote down! Could you please elaborate?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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