MHB Question about Linear Dependency

  • Thread starter Thread starter Yankel
  • Start date Start date
  • Tags Tags
    Linear
Click For Summary
The discussion centers on determining the linear dependence or independence of the vectors u1, v1, and w1 derived from the linearly dependent vectors u, v, and w. It is concluded that since u, v, and w are dependent, the new vectors u1, v1, and w1 must also be dependent, as they are expressed as linear combinations of u, v, and w. The coefficients of the linear combination indicate that at least one must be non-zero, reinforcing the dependence of u1, v1, and w1. The analysis suggests that the value of the scalar 'a' does not affect this conclusion. Therefore, the correct statement is that the vectors u1, v1, and w1 are linearly dependent for every value of a.
Yankel
Messages
390
Reaction score
0
which one of the next statements is the correct one ?

Let v,u,w be linearly dependent vectors in a vector space over R.

u1 = 2u
v1 = -3u+4v
w1 = u+2v-aw (a scalar from R)

(1) the vectors u1, v1 and w1 are linearly dependent for every value of a
(2) the vectors u1, v1 and w1 are linearly independent for every value of a
(3) the vectors u1, v1 and w1 are linearly independent for every value of a apart from 0
(4) the vectors u1, v1 and w1 are linearly independent for every positive value of a
(5) there exists a value of a for which the vectors u1, v1 and w1 are linearly independent

Thanks a lot !
 
Physics news on Phys.org
Yankel said:
which one of the next statements is the correct one ?

Let v,u,w be linearly dependent vectors in a vector space over R.

u1 = 2u
v1 = -3u+4v
w1 = u+2v-aw (a scalar from R)

(1) the vectors u1, v1 and w1 are linearly dependent for every value of a
(2) the vectors u1, v1 and w1 are linearly independent for every value of a
(3) the vectors u1, v1 and w1 are linearly independent for every value of a apart from 0
(4) the vectors u1, v1 and w1 are linearly independent for every positive value of a
(5) there exists a value of a for which the vectors u1, v1 and w1 are linearly independent

Thanks a lot !

This is somewhat similar to the last one. Again, we know that $u,v,w$ are dependent, implying that there are constants $c_1,c_2,c_3$ not all zero such that $c_1u+c_2v+c_3w=0$. Now, we want to analyze when the following is true:

\[d_1u_1+d_2v_1+d_3w_1=0\]

where $d_1,d_2,d_3\in\mathbb{R}$ are arbitrary constants. The idea now is to express the above equation in terms of a linear combination of just $u,v,w$, then use the fact that $u,v,w$ are linearly dependent to come up with the appropriate conclusion.

I hope this helps!
 
right, so if I am not mistaken I get:

(2d1-3d2+d3)u + (-4d2+2d3)v + (-ad3)w = 0

what does it tells me ? I know that at least one of the coefficients is not zero, because u,v and w are dependent...what can I say about u1,v1,w1 and what about a ?
 
i think focusing on the coefficients overmuch is a mistake.

it is clear that:

$\{u_1,v_1,w_1\} \subset \text{Span}(\{u,v,w\})$.

since {u,v,w} is linearly dependent, this has dimension ≤ 2.

therefore $u_1,v_1,w_1$ cannot be linearly independent, else we have a subspace of greater dimension than a space which contains it.

("a" is a red herring).
 
Thread 'How to define a vector field?'
Hello! In one book I saw that function ##V## of 3 variables ##V_x, V_y, V_z## (vector field in 3D) can be decomposed in a Taylor series without higher-order terms (partial derivative of second power and higher) at point ##(0,0,0)## such way: I think so: higher-order terms can be neglected because partial derivative of second power and higher are equal to 0. Is this true? And how to define vector field correctly for this case? (In the book I found nothing and my attempt was wrong...

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 59 ·
2
Replies
59
Views
9K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 5 ·
Replies
5
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K