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wurth_skidder_23
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Assume that [tex] m<n [/tex] and [tex] l_1,l_2,...,l_m [/tex] are linear functionals on an n-dimensional vector space
[tex] X [/tex].
Prove there exists a nonzero vector [tex] x [/tex] [tex] \epsilon [/tex] [tex] X [/tex] such that [tex] < x,l_j >=0 [/tex] for [tex] 1 \leq j \leq m[/tex]. What does this say about the solution of systems of linear equations?This implies
[tex] l_j(x) [/tex] [tex] \epsilon [/tex] [tex] X^\bot [/tex] for [tex] 1 \leq j \leq m [/tex] or [tex] l_j(x)=0 [/tex] for [tex] 1 \leq j \leq m [/tex]. Since it is stated in the problem that [tex] l_1,l_2,...,l_m [/tex] are linear functionals on the vector space X, [tex] l_j(x)=0 [/tex]. Does this reasoning even help me find the proof? I am stuck.
If you have trouble reading this, it is also at http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/olatex_33882.pdf
[tex] X [/tex].
Prove there exists a nonzero vector [tex] x [/tex] [tex] \epsilon [/tex] [tex] X [/tex] such that [tex] < x,l_j >=0 [/tex] for [tex] 1 \leq j \leq m[/tex]. What does this say about the solution of systems of linear equations?This implies
[tex] l_j(x) [/tex] [tex] \epsilon [/tex] [tex] X^\bot [/tex] for [tex] 1 \leq j \leq m [/tex] or [tex] l_j(x)=0 [/tex] for [tex] 1 \leq j \leq m [/tex]. Since it is stated in the problem that [tex] l_1,l_2,...,l_m [/tex] are linear functionals on the vector space X, [tex] l_j(x)=0 [/tex]. Does this reasoning even help me find the proof? I am stuck.
If you have trouble reading this, it is also at http://nirvana.informatik.uni-halle.de/~thuering/php/latex-online/olatex_33882.pdf
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