- #1

- 1,331

- 40

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter 1MileCrash
- Start date

- #1

- 1,331

- 40

- #2

- 144

- 1

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

- #3

fluidistic

Gold Member

- 3,767

- 134

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

- #4

- 1,331

- 40

Yes. Not all theorems work both ways like that.

Are you sure?

- #5

- 144

- 1

Are you sure?

I misread it, thought that the corollary was about dependance too.

- #6

- 1,331

- 40

I misread it, thought that the corollary was about dependance too.

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!)

Last edited:

- #7

Office_Shredder

Staff Emeritus

Science Advisor

Gold Member

- 4,435

- 518

- #8

- 1,331

- 40

_{2}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.

- #9

AlephZero

Science Advisor

Homework Helper

- 6,994

- 293

- #10

FlexGunship

Gold Member

- 369

- 8

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

Wait. Why? Because there's two extra commas in there?

Share: