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Linear Projection

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.

Solution

Let \dim_{\mathbb{R}} \mathcal{V} = n. Since any non zero vector is an eigenvector it follows that every basis of \mathcal{V} is also an eigenbasis. Let {\cal{B}}=\bigl\{{\overrightarrow{e_{i}}}\bigr\}_{i=1}^{n} be such a basis and \lambda_i \;, \; i=1, 2, \dots, n be the respective , not necessarily distinct , eigenvalues of the eigenvectors of \mathcal{B}. For the vector \overrightarrow{y}=\mathop{\sum}\limits_{i=1}^{n}\overrightarrow{e_{i}} which also happens to be eigenvector with eigenvalue \lambda , it holds that:

    \begin{align*} f\left ( \vec{y} \right ) = \lambda \vec{y} &\Rightarrow f\left ( \sum_{i=1}^{n} \overrightarrow{e_i} \right ) = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\!\!\!\!\!\!\!\!\overset{f \; \text{linear}}{=\! =\! =\! =\! =\!\Rightarrow } \sum_{i=1}^{n} f \left ( \overrightarrow{e_i} \right ) = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\Rightarrow \sum_{i=1}^{n} \lambda_i \overrightarrow{e_i} = \lambda \sum_{i=1}^{n} \overrightarrow{e_i} \\ &\Rightarrow \sum_{i=1}^{n} \left ( \lambda_i - \lambda \right ) \overrightarrow{e_i} =0 \\ &\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\overset{e_i \;\; \text{linearly independent}}{=\! =\! =\! =\! =\! =\!=\! =\!=\!=\!=\!=\!=\!\Rightarrow } \left [ \left ( \forall i=1, 2, \dots, n \right ) \;\; \lambda_i - \lambda =0 \right ] \\ &\Rightarrow \left [ \left ( \forall i=1, 2, \dots, n \right ) \;\; \lambda_i = \lambda \right ] \end{align*}

But then for each \overrightarrow{x}=\mathop{\sum}\limits_{i=1}^{n}x_i\overrightarrow{e_{i}}\in{\cal{V}} it holds that

    \[f\bigl({\overrightarrow{x}}\bigr)=f\left({\sum_{i=1}^{n}x_i\overrightarrow{e_{i}}}\right)=\displaystyle\sum_{i=1}^{n}x_if\bigl({\overrightarrow{e_{i}}}\bigr)=\sum_{i=1}^{n}x_i\lambda\,\overrightarrow{e_{i}}=\lambda\sum_{i=1}^{n}x_i\overrightarrow{e_{i}}=\lambda\overrightarrow{x}\]

The result follows.

Remark: This proof also works in the case \mathcal{V} is infinite.

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