Self-adjoint linear map on Euclidean space

[Grigorios Kostakos]

Let \(\left({{\cal{V}},\langle{\,,\,}\rangle}\right)\) a finite dimensional Euclidean vector space over \(\mathbb{R}\) (a vector space with inner product) and \(f:{\cal{V}}\longrightarrow{\cal{V}}\) self-adjoint linear transformation. Prove that if exists \(m\in{\mathbb{N}}\) such that\(f^m\equiv{\it{0}}\), then \(f\equiv{\it{0}}\), where \[f^m=\displaystyle\underbrace{f\circ{f}\circ\ldots\circ{f}}_{m-{\text{times}}}\,.\]

[Grigorios Kostakos]

By Spectral theorem for the self-adjoint linear map \(f:{\cal{V}}\longrightarrow{\cal{V}}\) we have that there exists orthonormal base \(\bigl\{{\overrightarrow{\varepsilon_1}\,, \ \overrightarrow{\varepsilon_2}\,,\ldots,\, \ \overrightarrow{\varepsilon_n}}\bigr\}\) of eigenvectors of \(f\).
So, for every \(i=1,2,\ldots,n\), exists an eigenvalue \(\lambda_i\), such that \[f\bigl({\overrightarrow{\varepsilon_i}}\bigr)=\lambda_i\,\overrightarrow{\varepsilon_i}\quad(1)\,.\] It can be proved inductively that if \(\lambda\in{\mathbb{R}}\) is an eigenvalue and \(\overrightarrow{x}\) the corresponding eigenvector of an endomorphism \(f :{\cal{V}}\longrightarrow{\cal{V}}\), then for every \(m\in{\mathbb{N}}\) holds: \[f^{m}\bigl({\overrightarrow{x}}\bigr)=\lambda^{m}\,\overrightarrow{x}\] and \(\lambda^{m}\) is an eigenvalue of endomorphism \(f^m\).
So, from \((1)\) we have that for every \(m\in{\mathbb{N}}\) holds \[f^{m}\bigl({\overrightarrow{\varepsilon_i}}\bigr)=\lambda_i^{m}\,\overrightarrow{\varepsilon_i}\quad(2)\] for \(i=1,2,\ldots,n \).
If for some \(m\in{\mathbb{N}}\) holds \(f^m={\it{0}}\), then for every \(i=1,2,\ldots,n\) we have \[{\it{0}}=f^{m}\bigl({\overrightarrow{\varepsilon_i}}\bigr)\stackrel{(2)}{=\!=}\lambda_i^{m}\,\overrightarrow{\varepsilon_i}\quad\stackrel{\overrightarrow{\varepsilon_i}\neq\overrightarrow{0}}{=\!\Longrightarrow}\quad\lambda_i^{m}=0\quad\Longrightarrow\quad\lambda_i=0\,.\] So the unique eigenvalue of \(f\) is \(0\) with multiplicity \(n\). But then, for every \(i=1,2,\ldots,n\) we have \[f\bigl({\overrightarrow{\varepsilon_i}}\bigr)=0\,\overrightarrow{\varepsilon_i}=\overrightarrow{0}\quad(3)\,.\] If \(\overrightarrow{y}=\mathop{\sum}\limits_{i=1}^{n}y_i\,\overrightarrow{\varepsilon_i}\in{\cal{V}}\), then \[\textstyle f\bigl({\overrightarrow{y}}\bigr)=f\bigl({\sum_{i=1}^{n}y_i\,\overrightarrow{\varepsilon_i}}\bigr)=\displaystyle\mathop{\sum}\limits_{i=1}^{n}y_i\,f\bigl({\overrightarrow{\varepsilon_i}}\bigr)\stackrel{(3)}{=\!=}\mathop{\sum}\limits_{i=1}^{n}y_i\,\overrightarrow{0}=\overrightarrow{0}\] and \(f={\it{0}}\).