Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
shuchii
Group Title
how do you find out if a vector is linearly dependent or not?
 one year ago
 one year ago
shuchii Group Title
how do you find out if a vector is linearly dependent or not?
 one year ago
 one year ago

This Question is Open

matricked Group TitleBest 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

shuchii Group TitleBest 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

KingGeorge Group TitleBest 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

shuchii Group TitleBest 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

KingGeorge Group TitleBest 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

KingGeorge Group TitleBest 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
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.