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matrickedBest ResponseYou've already chosen the best response.0
so for vectors v1,v2,v3,v4,........vn there exists scalars s1,s2,s3,s4......sn such that s1*v1 +s2*v2+ s3*v3 +.......................................sn*vn=0 then vectors v1,v2 ..... vn are linearly independent if each of s1,s22,s3...sn is 0 and linearly dependent if at least one of s1,s22,s3...sn is not 0
 one year ago

shuchiiBest ResponseYou've already chosen the best response.0
soo if my vectors are [0 2 3], [0 0 8], and [1 3 1], they would be independent?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
I believe those would be independent. Suppose that there are some scalars \(c_1,c_2\in\mathbb{R}\) such that \[c_1\cdot[0,2,3]+c_2[0,0,8]=[1,3,1]\]So\[[0,2c_1,3c_1]+[0,0,8c_2]=[1,3,1]\]However, \(0+0=0\neq1\) for any choice of \(c_1\) and \(c_2\). You can make similar arguments to show that \([0,2,3]\) and \([0,0,8]\) are similarly independent. This would show that all three are linearly independent.
 one year ago

shuchiiBest ResponseYou've already chosen the best response.0
woahhh that makes so much sense! [: thank you :) can you make that argument all the time?
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
You can try to make that argument. If you're working in \(\mathbb{R}\) or the complex numbers, it's usually fairly straightforward (at least in what you will be expected to do). An alternative method, which works particularly well when you have 3 or 2dimensional vectors, is to put them in a matrix. \[\begin{bmatrix}1&3&1\\0&2&3\\0&0&8\end{bmatrix}\]If you can rowreduce to this form, they're linearly independent.
 one year ago

KingGeorgeBest ResponseYou've already chosen the best response.1
Another thing to watch for, is the dimension of the vector space. If you have an \(n\) dimensional vector space, and \(m\) vectors, where \(m>n\), then your vectors are not linearly independent.
 one year ago
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