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Romero
 2 years ago
I remember reading that if you have a set of vectors that if one of the vectors is a lin combination of the vectors before it then the set is lin dependent. What I'm confused about is does that vector have to be lin combination of ALL of the previous vectors or just at least one for the set to be lin dependent?
Romero
 2 years ago
I remember reading that if you have a set of vectors that if one of the vectors is a lin combination of the vectors before it then the set is lin dependent. What I'm confused about is does that vector have to be lin combination of ALL of the previous vectors or just at least one for the set to be lin dependent?

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Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1a combination of one or more though if you can write it is a l.c. of one of the vectors you can write it is a linear combination of all the vectors

glgan1
 2 years ago
Best ResponseYou've already chosen the best response.0I think should be all the combinations.

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Great so if you want to get the basis of the set of vectors you get rid of the linear combination right?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1you get rid off was many vectors as needed to make them independent (but no more)

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Ok let's say we have a a set of 3 vectors. We can't determine if it's lin dependent by inspection yet when we reduce it to echelon form we find a free variable so we only have two pivot points but this was done through row reduction using all the rows meaning we added or subtracted row 1 to row 2 and then used row 2 to make the third row a zero row. At that point if we want to have the basis of the set we have to get rid of a vector. Can we get rid of any vector?

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Can we get rid of any vector? @Zarkon This is basically what I'm asking :)

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0That's weird. So you can have three different basis for the set of vectors?

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Does this mean that basis is not unique?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1correct the basis is not unique...there can be an infinite number of basis's

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0Is there a condition where a basis will be unique?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1dw:1336751060283:dwdw:1336751083709:dw both a basis for \(\mathbb{R}^2\)

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1you would have to be pretty restrictive

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0What a bais for a vector space? Can that be infinite as well?

Romero
 2 years ago
Best ResponseYou've already chosen the best response.0and you can do this by multiplying the basis by a scalar?

Zarkon
 2 years ago
Best ResponseYou've already chosen the best response.1it is really too complicated to talk about here. I'd recommended you read http://en.wikipedia.org/wiki/Banach_space
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