Am I the only one that finds this funny?

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Is this really worthy of being pointed out as a corollary??

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Oops. It's just the contrapositive. Then no, it's not worthy of being a corollary.

fluidistic
Gold Member
If it follows from the theorem 1.6 then why not? :) Not really funny to me.

Yes. Not all theorems work both ways like that.
Are you sure?

Are you sure?

Hah, I don't blame you. I had to do a double (or triple) take.

Though, interestingly enough, if the corollary was about dependence, then that would mean (S1 dependent) <=> (S2 dependent), which would mean that any subset or superset of any linearly dependent set would be linearly dependent. Which would mean that all subsets of a vector space would be linearly dependent, because all vector spaces must contain some linearly dependent subset (namely itself) (unless I'm just confusing myself here!)

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Office_Shredder
Staff Emeritus
Gold Member
That's true, but not for the reason that you seem to be thinking. Every vector space contains 0, and therefore is a linearly dependent set, but if you remove that point you could have a linearly independent set. For example (and probably the only example), F2 the field of two elements, as a one dimensional vector space over itself.

That's true, but not for the reason that you seem to be thinking. Every vector space contains 0, and therefore is a linearly dependent set, but if you remove that point you could have a linearly independent set. For example (and probably the only example), F2 the field of two elements, as a one dimensional vector space over itself.
I'm saying - if the corollary was talking about dependence, a consequence would be that any subset of a vector space is linearly dependent because the vector space itself is.

AlephZero