Question

$Daca~A,~B\in~M_n(R)~si~AB=BA~atunci~det(A^2+B^2)\geq 0.$

Answer 1

$\displaystyle Clasica~si~frumoasa! \\ \\ Cum~A~si~B~comuta,~rezulta~A^2+B^2=(A+iB)(A-iB). \\ \\ Fie~P(x)=\det(A+xB).~P(x)~este~un~polinom,~iar~pentru~ca \\ \\ A,B \in \mathcal{M}_n(\mathbb{R}),~rezulta~P(x) \in \mathbb{R}[x]. \\ \\ Atunci~P(\overline{z})= \overline{P(z)}~\forall~z \in \mathbb{C}. \\ \\ Deci~P(-i)= \overline{P(i)}.$

$\displaystyle \det(A^2+B^2)= \det(A+iB)(\det(A-iB)=P(i)P(-i)= \\ \\ =P(i) \overline{P(i)}=|P(i)|^2\ge 0,~Q.E.D.$