Linear Dependence of Vectors in R3: Exploring Why S is Not a Basis

[will]

Homework Statement

4) explain why S is not a basis for R^3
(a) S={(1,2,-1),(0,0,0),(1,0,1)}
(b) explain why S is not a basis for R^3
S={(2,4,5),(-1,3,6),(7,7,9),(-4,2,-4)}

Homework Equations

all proofs

The Attempt at a Solution

(a)**The set S is a set with the 0 vector, (0,0,0). Such a set is always dependant and can therefore never be a basis for R3.

The set without the zero vector S={(1,2,-1),(1,0,1} is also no basis for R3, it contains less vectors than the dimension of R3 (which is 3).

(b)**To have a bas you need only 3 vectors by having 4 that makes this system linearly dependent and there for it doesn't represent a base.. to make it a base one has to be eliminated and check if they have linear independency


"" is there a better more elegant answer, by justifying an Axion or a theorem" my teacher is so hard" thanks guys

 

[verafloyd]

Ahh, teachers :)

Okay, search for what "rank" means if you already don't know, then relate the number of linearly independent vectors in S to the rank of S (smaller, larger, equal...). This would be the elegant way of explaining it :)

 

[radou]

"" is there a better more elegant answer, by justifying an Axion or a theorem" my teacher is so hard" thanks guys

The answers are elegant enough, it's just using definitions of "basis", "span", etc.

 

[HallsofIvy]

Did your teacher ask for a "more elegant" way? In each case you have said that the set of vectors is not a basis because it is not linearly independent. That's showing that the definition of "base" is not satisfied and is plenty "elegant".
(Of course, you do not need the other comments on "The set without the zero vector" and " to make it a base one has to be eliminated and check if they have linear independency" because those questions were not asked.)