Ask your own question, for FREE!
MIT 18.06 Linear Algebra, Spring 2010 11 Online
OpenStudy (kenshin):

Hi I need some help remembering the definition and conditions of a basis: In which way can the following implications go? 1) => 2), 2) => 1) or both? 1) A basis for a subspace or any space has to satisfy the following 2 conditions: a) be linearly independent, b) span the subspace; 2) The "Basis theorem 15" in David C. Lay's 3rd edition of Linear Algebra and its applications says: Let H be a p-dimensional subspace of R^n. Any linearly independent set of exactly p elements in H is automatically a basis for H. Also, any set of p elements of H that spans H is automatically a basis for H.

OpenStudy (kenshin):

I'm confused because: 2) seems to suggest that condition a) implies condition b) for whichever it satifies, the other must also be true. but in 1) it seems to suggest that both a) and b) need to be satisfied before we can even conclude it being a basis to begin with.

OpenStudy (anonymous):

basis meaning number of independent columns mean to say columns having pivots in matrix are basis of the matrix

OpenStudy (phi):

Any linearly independent set of exactly p elements in H notice they say "exactly p elements" Condition 1 gets to the same thing by saying a) be linearly independent, b) span the subspace; the "span" part means you need exactly p elements (for a subspace with p dimensions) without part (b) , you could have less than p independent elements i.e. not enough. if they just said (b), you could have more than p elements i.e. some would be linearly dependent. In other words, spanning implies you have at least p elements, but it might mean more than p (and therefore some redundant bases)

OpenStudy (kenshin):

Thanks!

OpenStudy (banda_mohammod_al_helal):

in\[R^n\] at most n vectors/equation/colums can be independent and for \[R^n\] there must n vectors to span/cover the full space.the basis are those vectors that are linearly independent and span the space.So,n independent vectors are basis of \[R^n\]

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!