Azi, la seminarul de la Institut de algebre de operatori, domnul profesor Stratila a facut o problema foarte interesanta, anume http://mateforum.ro/viewtopic.php?p=678#678
Aici am sa postez un fel de "echivalent" al problemei din link-ul de mai sus la
clasa 11-a. Iata problema:
Sa se arate ca pentru orice doua matrice \( A, B\in M_{2}(\mathbb{R}) \) are loc inegalitatea:
\( \det(A^2+B^2)\geq\det(AB-BA) \).
Cand are loc egalitatea?
det(A^2+B^2)\geq det(AB-BA)
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Cezar Lupu
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det(A^2+B^2)\geq det(AB-BA)
Last edited by Cezar Lupu on Tue Oct 16, 2007 9:26 am, edited 1 time in total.
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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Cezar Lupu
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Totusi, hai sa dam o indicatie:
Scrieti \( (A+iB)(A-iB)=A^{2}+B^{2}+i(BA-AB) \) si \( (A-iB)(A+iB)=A^{2}+B^{2}+i(AB-BA) \), iar \( \det(A+iB)(A-iB)=\det(A-iB)(A+iB)=\det(A^{2}+B^{2}) \).
Scrieti \( (A+iB)(A-iB)=A^{2}+B^{2}+i(BA-AB) \) si \( (A-iB)(A+iB)=A^{2}+B^{2}+i(AB-BA) \), iar \( \det(A+iB)(A-iB)=\det(A-iB)(A+iB)=\det(A^{2}+B^{2}) \).
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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Radu Titiu
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Fie \( P(X)=\det(A^2+B^2-X(AB-BA))=aX^2+bX+c \)avand coeficientii \( a,b,c \in \mathbb{R} \), \( a=\det(AB-BA) \), \( c=\det(A^2+B^2). \)
Astfel avem:
\( \det(P(i))=\det(A+iB)\det(A-iB)=|\det(A+iB)|^2 \geq 0 \)
Deoarece \( P(i) \in \mathbb{R}_{+} \Rightarrow b=0 \) , astfel \( c-a \geq 0 \) ceea ce trebuia demonstrat.
Astfel avem:
\( \det(P(i))=\det(A+iB)\det(A-iB)=|\det(A+iB)|^2 \geq 0 \)
Deoarece \( P(i) \in \mathbb{R}_{+} \Rightarrow b=0 \) , astfel \( c-a \geq 0 \) ceea ce trebuia demonstrat.
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Bogdan Posa
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frumoasa solutie Svejk...
\( \det [A^{2}+B^{2}+i(AB-BA)]+\det [A^{2}+B^{2}-i(AB-BA)] \)=
\( \det (A+iB)(A-iB)+\det (A-iB)(A+iB) \) \( \geq 0 \)
Dar
\( \det [A^{2}+B^{2}+i(AB-BA)]+\det [A^{2}+B^{2}-i(AB-BA)] \)=
\( 2 \det (A^{2}+B^{2})+2 \det [i(AB-BA)] \)=
\( 2\det (A^{2}+B^{2})-2\det (AB-BA)\geq 0 \).
\( \det [A^{2}+B^{2}+i(AB-BA)]+\det [A^{2}+B^{2}-i(AB-BA)] \)=
\( \det (A+iB)(A-iB)+\det (A-iB)(A+iB) \) \( \geq 0 \)
Dar
\( \det [A^{2}+B^{2}+i(AB-BA)]+\det [A^{2}+B^{2}-i(AB-BA)] \)=
\( 2 \det (A^{2}+B^{2})+2 \det [i(AB-BA)] \)=
\( 2\det (A^{2}+B^{2})-2\det (AB-BA)\geq 0 \).