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Fie \( A\in\mathcal{M}_n(\mathbb{R}) \) astfel incat \( A^{T}=A^{*} \). Demonstrati ca daca \( S(A) \) este suma tuturor elementelor matricei \( A \), atunci:

\( (S(A))^2\leq n^3 \cdot \det A \)

Folosind relatia \( A\cdot A^{\ast}=(\det A) I_n \) obtinem \( A\cdot A^T=(\det A) I_n \), deci \( \sum_j {a_{ij}^2}=\det A \) pentru orice \( i \).

Apoi aplicand CBS se obtine concluzia.

\( LHS=(\sum _{i,j}{a_{ij})^2\le n\sum_i{(\sum_j{a_{ij})^2}}}\le n^2\sum_i{\sum_j{a_{ij}^2}}=RHS \)