## shuchii 2 years ago how do you find out if a vector is linearly dependent or not?

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1. matricked

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

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

3. KingGeorge

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

woahhh that makes so much sense! [: thank you :) can you make that argument all the time?

5. KingGeorge

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

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.