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shuchii

  • 2 years ago

how do you find out if a vector is linearly dependent or not?

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  1. matricked
    • 2 years ago
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    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

  2. shuchii
    • 2 years ago
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    soo if my vectors are [0 2 3], [0 0 -8], and [1 -3 1], they would be independent?

  3. KingGeorge
    • 2 years ago
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    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.

  4. shuchii
    • 2 years ago
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    woahhh that makes so much sense! [: thank you :) can you make that argument all the time?

  5. KingGeorge
    • 2 years ago
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    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.

  6. KingGeorge
    • 2 years ago
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    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.

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