## anonymous 3 years ago Concept-checking 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($$\vec{u}$$, $$\vec{v}$$) ?

1. anonymous

(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.

2. anonymous

If 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?

3. anonymous

I believe so, yes...

4. UnkleRhaukus

yes yes yes.

5. anonymous

And so, if they do not span $$\Re^3$$, they are linearly dependent, right?

6. UnkleRhaukus

yeah three vectors that aren't all linearly independent cannot span $\mathbb R^3$

7. anonymous

Could "yeah three vectors that aren't all linearly independent cannot span $$\Re^3$$" be the justification of the answer?

8. UnkleRhaukus

i wouldn't word it like that

9. anonymous

Since 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?

10. UnkleRhaukus

thats is better yes

11. anonymous

That's pretty nice :) Thanks! May I ask one more question about linear dependence here?

12. UnkleRhaukus

sure

13. anonymous

Checking 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?

14. UnkleRhaukus

for 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

15. anonymous

I 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!! :)

16. anonymous

If $$\vec{w}$$ can be written as a linear combination of $$\vec{u}, \vec{w}$$ then they are dependent.

17. anonymous

$$\vec{v}$$ **

18. anonymous

And 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" :|

19. UnkleRhaukus

if you find the angle between vectors is π/2 the vectors will be independent

20. anonymous

How to find the angle between three vectors??

21. UnkleRhaukus

$\theta=\arccos\left(\frac{\vec u \cdot \vec v}{||\vec u||~~||\vec v||}\right)$

22. UnkleRhaukus

For multiple vectors you would have to find the angle between each possible pair of vectors

23. anonymous

For the case of two vectors, I still can do it. But not for three or more vectors..