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Mathematics 23 Online
OpenStudy (anonymous):

Every nonzero vector space V contains a nonzero proper subspace true false

OpenStudy (anonymous):

please explain your ans

OpenStudy (zarkon):

false

OpenStudy (anonymous):

how false

OpenStudy (zarkon):

look at the vector space that has the basis vector \{(1,0)\} so V=span( \{(1,0)\}) the only proper subspace of V is the zero subspace

OpenStudy (zarkon):

\[V=\text{span}(\{(1,0)\})\]

OpenStudy (zarkon):

do you see why?

OpenStudy (anonymous):

plz explain if u can ''the only proper subspace of V is the zero subspace''

OpenStudy (zarkon):

Suppose W is a proper subspace of V and that W is not the zero subspace....

OpenStudy (zarkon):

then there is a nonzero vector \(w\in W\)

OpenStudy (zarkon):

the vector \(w\) has the form (x,0) where \(x\ne 0\) then \[(x,0)/x=(1,0)\in W\] thus W=V ...a contradiction

OpenStudy (anonymous):

what is proper subspace? is it like proper subset

OpenStudy (zarkon):

sort of

OpenStudy (zarkon):

W is a proper subspace of V if W is a vector space and every vector in W is in V, but W\(ne\)V

OpenStudy (zarkon):

W\(\ne\)V

OpenStudy (anonymous):

means proper subspace is actually a proper subset in which properties of vector space is satistied right?

OpenStudy (zarkon):

yes

OpenStudy (anonymous):

ok thanks a lot

OpenStudy (zarkon):

np

OpenStudy (anonymous):

@zarkon please help if u can. Similar matrices represent the same linear transformation. please prove it or send me a link of its prove its urgent....

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