Ivan Shishkin, Rye (1878)

Problems/Linear algebraUnreviewed

Valeurs propres de matrices orthogonales

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Soit AA une matrice orthogonale. Que peut-on dire du moduleEN des valeurs propres de la matrice AA ?

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Solution by AndurilEN

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The only possible modulus is 1.

Indeed, an eigenvalue is bounded above by any sub-multiplicative norm. But the subordinate to the euclidean norm of such a matrix is 1 so we have that |a|<=1 for such an eigenvalue. Yet, 1/a is an eigenvalue of the inverse of the orthogonal matrix, which is also orthogonal, hence |1/a|<=1 and |a|>=1.

Conversely, the following matrix is orthogonal and has e^ix as an eigenvalue :
(cosxsinxsinxcosx)\begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix}

Solution by visitor

Discussions0 useful votes

Toutes les valeurs propres complexes d’une matrice orthogonale sont de module 11.

Soit AOn(R)A\in\mathcal{O}_n(\mathbb{R}), c’est-à-dire ATA=InA^{\mathsf T}A=I_n. On travaille dans Cn\mathbb{C}^{n}, muni du produit hermitien x,y=xTy\langle x,y\rangle=\overline{x}^{\mathsf T}y.
Soit λC\lambda\in\mathbb{C} une valeur propre de AA et xCn{0}x\in\mathbb{C}^{n}\setminus\{0\} un vecteur propre associé. Alors
x2=xTx=xTAAx=(Ax)T(Ax)=Ax2=λx2=λ2x2.\|x\|^{2}=\overline{x}^{\mathsf T}x=\overline{x}^{\mathsf T}A^{*}Ax=\overline{(Ax)}^{\mathsf T}(Ax)=\|Ax\|^{2}=\|\lambda x\|^{2}=|\lambda|^{2}\|x\|^{2}.Comme x0x\neq 0, on a x2>0\|x\|^{2}>0 et l’on peut simplifier. Ainsi λ2=1|\lambda|^{2}=1, donc λ=1|\lambda|=1.

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