A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years ago
Conceptchecking questions
Consider vectors \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\), if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\), does that mean that
i) \(\vec{w}\) is linearly dependent on \(\vec{u}\) and \(\vec{v}\) ?
ii) sp(\(\vec{u}\), \(\vec{v}\), \(\vec{w}\)) = sp(\(\vec{u}\), \(\vec{v}\)) ?
anonymous
 3 years ago
Conceptchecking questions Consider vectors \(\vec{u}\), \(\vec{v}\) and \(\vec{w}\), if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\), does that mean that i) \(\vec{w}\) is linearly dependent on \(\vec{u}\) and \(\vec{v}\) ? ii) sp(\(\vec{u}\), \(\vec{v}\), \(\vec{w}\)) = sp(\(\vec{u}\), \(\vec{v}\)) ?

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0(i) follows straight from the definition of linear dependence :) For (ii), recognize that the linear span of a set of vectors is the set of all its finite linear combinations; if \(\vec{u},\vec{v},\vec{w}\) are linearly dependent, i.e. \(\vec{w}\) is a linear combination of \(\vec{u},\vec{v}\), then it should be immediately apparent that the spanning sets are identical.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If I can show \(\vec{u}=(u_1, u_2, u_3)\), \(\vec{v}=(v_1, v_2, v_3)\), and \(\vec{w}=(w_1, w_2, w_3)\) span \(\Re^3\), can I immediately draw a conclusion that the vectors are linearly independent?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And so, if they do not span \(\Re^3\), they are linearly dependent, right?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1yeah three vectors that aren't all linearly independent cannot span \[\mathbb R^3\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Could "yeah three vectors that aren't all linearly independent cannot span \(\Re^3\)" be the justification of the answer?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1i wouldn't word it like that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Since it cannot span \(\Re^3\), that means one of the vectors is a linear combination of the other vectors, therefore, the vectors are linearly dependent. ^ Could that be one?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1thats is better yes

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0That's pretty nice :) Thanks! May I ask one more question about linear dependence here?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Checking the linear dependence of vectors, say, \(\vec{u} =(u_1,u_2,u_3), \vec{v} =(v_1,v_2,v_3),\) and \(\vec{w} =(w_1,w_2,w_3)\) Method I) \[\left[\begin{matrix}u_1 & v_1 &  & w_1 \\ u_2 & v_2 &  & w_2 \\u_3 & v_3 &  &w_3\end{matrix}\right]\] Solve it, if it is a) consistent (that is it has solution), then it is linearly dependent b) inconsistent (that is it has no solution), the it is linearly independent Method II) Find the determinant of \[\left[\begin{matrix}u_1 & v_1 & w_1 \\ u_2 & v_2 & w_2 \\u_3 & v_3 & w_3\end{matrix}\right]\] If the determinant is a) equal to 0, it is linearly dependent b) not equal to 0, it is linearly independent Is that correct? Is there any other way to check the linear dependence?

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1for the first method if (0,0,0) is the only solution the the vectors are independent the second method is right, and im sure there are other methods but i can think of them

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think for the first method, it's actually testing if \(\vec{w}\) is a linear combination of \(\vec{u}\) and \(\vec{v}\) Please leave a comment here if you can think of any of them. That would be really helpful! Thank you for your help!! :)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If \(\vec{w}\) can be written as a linear combination of \(\vec{u}, \vec{w}\) then they are dependent.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0And that's why "if it is a) consistent (that is it has solution), then it is linearly dependent b) inconsistent (that is it has no solution), the it is linearly independent" :

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1if you find the angle between vectors is π/2 the vectors will be independent

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0How to find the angle between three vectors??

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1\[\theta=\arccos\left(\frac{\vec u \cdot \vec v}{\vec u~~\vec v}\right)\]

UnkleRhaukus
 3 years ago
Best ResponseYou've already chosen the best response.1For multiple vectors you would have to find the angle between each possible pair of vectors

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0For the case of two vectors, I still can do it. But not for three or more vectors..
Ask your own question
Sign UpFind more explanations on OpenStudy
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.