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Fie \( A,B\in M_{2n+1}(\mathbb{R}) \) astfel incat \( A^2-B^2=I_{2n+1} \). Demonstrati ca \( \det(AB-BA)=0 \).

Mihai Opincariu G.M.B. 2008

Se arata usor ca \( AB^2A=BA^2B \), deci matricile AB si BA comuta.

De aici \( (AB-BA)^{2p+1}=(AB)^{2p+1}-C_{2p+1}^1(AB)^{2p}(BA)+...-(BA)^{2p+1} \) si aplicand urma obtinem \( \tr(AB-BA)^{2p+1}=0 \) pentru orice p natural.

De aici folosind Formulele lui Newton obtinem ca AB-BA are cel putin o valoare proprie nula, deci are determinantul 0.

Intr-adevar, daca \( t_1,t_2,...t_{2n+1} \) sunt valorile proprii ale matricei AB-BA, atunci:

\( \left{\begin{array}{cc}t_1+t_2+...+t_{2n+1}=0\\t_1^3+t_2^3+...+t_{2n+1}^3=0\\...............................................\\t_1^{2n+1}+t_2^{2n+1}+...+t_{2n+1}^{2n+1}=0\end{array} \)

Insa
1) \( s_p-s_{p-1}\sigma_1+s_{p-2}\sigma_2-...+(-1)^{p-1}s_1\sigma_{p-1}+(-1)^p\sigma_p=0 ,(\forall)p\in\mathbb{N^{\ast}},1\le p\le 2n+1 \)
si
2) \( s_p-s_{p-1}\sigma_1+s_{p-1}\sigma_2-...+(-1)^{2n+1}s_{p-2n-1}\sigma_{2n+1}=0 , (\forall)p\in\mathbb{N^{\ast}},p\ge 2n+2 \),

de unde succesiv

p=1: \( s_1-\sigma_1=0 \) de unde \( \sigma_1=s_1=0 \)

p=3: \( s_3-s_2\sigma_1+s_2\sigma_2-3\sigma_3=0 \) de unde \( \sigma_3=0 \)

p=5: \( s_5-s_4\sigma_1+s_3\sigma_2-s_2\sigma_3+s_2\sigma_4-5\sigma_5=0 \) de unde \( \sigma_5=0 \)

Recursiv obtinem \( \sigma_p=0 \) pentru orice p impar \( 1\le p\le 2n+1 \), deci \( t_1\cdot t_2\cdot ....\cdot t_{2n+1}=0 \).

Notatii: \( s_p=\sum t_1^p=\sum_{i=1}^{2n+1}t_i^p \) si \( \sigma_k=\sum t_1t_2...t_k=\sum_{1\le i_1< i_2<...\le n}t_{i_1}t_{i_2}...t_{i_k} \)

Pentru \( X=A-B , Y=A+B \) relatia din enunt se scrie \( XY-I=I-YX \) si cum se stie ca \( \det(XY-I)=\det(YX-I) \) obtinem \( \det(XY-I)=0 \) adica \( \det(AB-BA)=0 \) .