Fie \( A, B\in M_{2}(\mathbb{R}) \) doua matrice astfel incat \( AB=BA \). Sa se arate ca \( \det(A^{2}+B^{2})\geq(\det A- \det B) ^{2} \).
Tudorel Lupu, R.M.I. C-ta, 2003
A si B comuta, atunci det(A^2+B^2)>=(detA-detB)^2
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Cezar Lupu
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A si B comuta, atunci det(A^2+B^2)>=(detA-detB)^2
An infinite number of mathematicians walk into a bar. The first one orders a beer. The second orders half a beer. The third, a quarter of a beer. The bartender says “You’re all idiots”, and pours two beers.
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Marius Mainea
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Putem presupune ca A sau B inversabila, altfel e trivial.
Sa zicem ca B inversabila. ,,Fortand'' factor comun B e suficient sa demonstram relatia:
\( \det (X^2+I_2)\geq (\det X-1)^2 \)
Ori aceasta se reduce la
\( \det^2X+\tr X^2+1\geq \det^2X-2\det X+1 \)
sau
\( (\tr X)^2-2\det X\geq -2\det X \)
ceea ce este evident adevarat.
Sa zicem ca B inversabila. ,,Fortand'' factor comun B e suficient sa demonstram relatia:
\( \det (X^2+I_2)\geq (\det X-1)^2 \)
Ori aceasta se reduce la
\( \det^2X+\tr X^2+1\geq \det^2X-2\det X+1 \)
sau
\( (\tr X)^2-2\det X\geq -2\det X \)
ceea ce este evident adevarat.