Pentru orice matrice \( X \in M_{n}(R) \) notam cu \( s(X) \) suma tuturor elementelor sale. Aratati ca pentru orice matrice \( A,B \in M_{n}(R) \) avem: \( s(AB^{T})s(BA^{T}) \leq s(AA^{T})s(BB^{T}) \), unde \( X^{T} \) este transpusa matricei \( X \).
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Suma elementelor dintr-o matrice
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bogdanl_yex
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Suma elementelor dintr-o matrice
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Beniamin Bogosel
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Daca facem calculele pentru \( X=(x_{ij}) \) si \( A=(a_{ij}),\ B=(b_{ij}) \) avem
\( s(XX^T)=\sum\limits_{i=1}^n(x_{1i}+...+x_{ni})^2 \)
\( s(AB^T)=\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})(b_{1i}+...+b_{ni}) \)
Evident \( s(X)=s(X^T) \) deci \( s(AB^T)=s(BA^T) \).
Astfel inegalitatea devine
\( s(AB^T)^2\leq s(AA^T)s(BB^T)\Leftrightarrow \left(\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})(b_{1i}+...+b_{ni})\right)^2\leq (\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})^2)(\sum\limits_{i=1}^n(b_{1i}+...+b_{ni})^2) \)
care este inegalitatea Cauchy-Buniakowski-Schwarz.
\( s(XX^T)=\sum\limits_{i=1}^n(x_{1i}+...+x_{ni})^2 \)
\( s(AB^T)=\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})(b_{1i}+...+b_{ni}) \)
Evident \( s(X)=s(X^T) \) deci \( s(AB^T)=s(BA^T) \).
Astfel inegalitatea devine
\( s(AB^T)^2\leq s(AA^T)s(BB^T)\Leftrightarrow \left(\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})(b_{1i}+...+b_{ni})\right)^2\leq (\sum\limits_{i=1}^n(a_{1i}+...+a_{ni})^2)(\sum\limits_{i=1}^n(b_{1i}+...+b_{ni})^2) \)
care este inegalitatea Cauchy-Buniakowski-Schwarz.