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shuchii
how do you find out if a vector is linearly dependent or not?
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
soo if my vectors are [0 2 3], [0 0 -8], and [1 -3 1], they would be independent?
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.
woahhh that makes so much sense! [: thank you :) can you make that argument all the time?
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 2-dimensional vectors, is to put them in a matrix. \[\begin{bmatrix}1&-3&1\\0&2&3\\0&0&8\end{bmatrix}\]If you can row-reduce to this form, they're linearly independent.
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.