# Question about Linear Dependency

#### Yankel

##### Active member
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 !!

#### Chris L T521

##### Well-known member
Staff member
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!

#### Yankel

##### Active member
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 ?

#### Deveno

##### Well-known member
MHB Math Scholar
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).