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Anonymous Thu 29 Jul 12:32:43 2021 No.13448331 View Reply Original Report

Quoted By: >>13448333 >>13448344 >>13448889 >>13449945 >>13450128

Quick linear algebra question: when trying to determine if a group of vectors are linearly dependent, am I supposed to find a solution for the zero vector where NONE of the scalars are 0 or when just at least one is not 0

Anonymous

Anonymous Thu 29 Jul 2021 12:33:57 No.13448333 Report

Quoted By:

>>13448331
dunno

Anonymous

Anonymous Thu 29 Jul 2021 12:40:07 No.13448344 Report

Quoted By:

>>13448331
A group of vectors are linearly dependant when either the colums or the rows are multiples of each other. A vector is independent if it "add dimension" to the column space of a matrix.

Anonymous

Anonymous Thu 29 Jul 2021 15:49:33 No.13448889 Report

Quoted By:

>>13448331
At least one needs to be non-zero, that is all.
The idea is that there is always a way to produce the zero vector for any group of vectors by setting all the scalars to zero, and so we exclude this case in particular. Any other case is good.

Anonymous

Anonymous Thu 29 Jul 2021 15:53:58 No.13448903 Report

Quoted By:

If they are linearly independent, then only the trivial solution exists (Ax=0, where x=0 vector). So if they are dependent, then at least one solution exists where x!=0 vector.

Anonymous

Anonymous Thu 29 Jul 2021 20:16:03 No.13449945 Report

Quoted By:

>>13448331
You need to find a linear combination where at least one of the coefficients is nonzero.

Linearly Independent : if there is a linear combination summing to zero, EVERY FUCKING COEFFICIENT IS ZERO

Linearly dependent: not linearly independent. Thus, there exists a linear combination summing to zero with at least one nonzero coefficient.

Remember, linear dependence is "redundancy". If you have 0v_1+5v_2+0v_3=0, then v_2=0 and hence redundant.

Anonymous

Anonymous Thu 29 Jul 2021 21:03:39 No.13450128 Report

Quoted By:

>>13448331
Think, bro, think.

Take any nonzero vector v. Consider the set {v}. Is it linearly independent? Yes, because there is only one vector (and it is not the zero vector). Now take the set {v, v}. Is it linearly independent? Heck no, because v - v = 0. Now take the set {v, v, v} Is it linearly independent. Heck no again, but we can also have 0v + v - v =0, so there is a zero coefficient, but the set is linearly dependent.

Anonymous

Anonymous Thu 29 Jul 2021 21:18:29 No.13450169 Report

Quoted By:

If there exists a nontrivial solution, the vectors are linearly dependent.

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