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Multiply each vector with C's and then set the linear combination of them equal to zero. Solve for each component. If you eventually find that the C's are nonzero, you can conclude that the vectors are dependent. If they equal zero, they are linearly independent. Depending on whether the vectors are solutions to a differential equation, you can also use the Wronskian.
Triple post, my bad.
the vector set they give us is [B1, B2,....Bm],
Do \[C_1B_1+C_2B_2+C_NB_N=0\] and factor where possible.
k thanks, do you see where this would be proven though?
nvm I think I read the question wrong
Are you figuring out subspaces?
the question is if the vectors B1,B2..Bn is linearly dependent, then any collection of vectors which contains these vectors is also linearly independent....which is easy
Show that two planar vectors alpha and beta are linearly independent if and only if they are not parallel
The idea is, the vectors are parallel, then they can be scalar multiples of eachother. That implies that \[C_1\] and \[C_2\] are constants, therefore they are dependent.
Show that three vectors a,b,y which lie in the same plane must be linearly dependent
Same idea as the last part. If vectors are colinear in the plane, then they must be dependent. If not, a linear combination of any two vectors can be colinear with the other, meaning that they are linearly dependent. Same logic, more gimmicks.
So if they are all in the same plane, they can be represented by some variation of (S1V1+S2V2=a)
k thanks again, how you feeling about your test tomorrow/
Pretty good. I'm not too worried, just gotta be able to derive something if I go absent minded.