Two linearly dependent vectors

1. Mar 6, 2009

bcjochim07

1. The problem statement, all variables and given/known data

Prove that two vectors are linearly dependent if and only if one is a scalar multiple of the other.

2. Relevant equations

3. The attempt at a solution This seems at glance to be a fairly easy proof:

Part I Assume that vectors u and v are linearly dependent.

Then c1u + c2v = 0 where c1 and c2 are not both 0

then u = -c2/c1 * v
and v = -c1/c2 * u But this doesn't make sense to me because what if one of c1 or c2 does equal zero?

Part II Assume that u =av and v=bu , where a and b are constants

then u - av = 0 where the coefficient of u is 1 and v - bu = 0 where the coefficient of v is 1.

Therefore u and v are linearly dependent.

I'm struggling a bit with linear algebra proofs, so any critique or suggestions that anyone could offer would be greatly appreciated.

2. Mar 6, 2009

phreak

Either c1 or c2 is nonzero. Without loss of generality, let c1 be nonzero. Then since c1u + c2v = 0, dividing by c1, we get u + (c2/c1)v = 0. What does this tell you?

3. Mar 6, 2009

gabbagabbahey

Look at two different cases:

Case1: c1=0 and c2≠0

Case 2: c2=0 and c1≠0

...your method in partII is not valid, since it neglects the c1=0 case.

4. Mar 6, 2009

bcjochim07

Ok so v = c1u

v -c1u = 0 and suppose that c1 = 0.
v must be the zero vector

now suppose that v-c1=0 and c1 is not equal to zero.
then v-c1u=0 where the coefficients of v and u are both nonzero

Am I going on the right track?

and then I would do a similar thing for u = c2v

5. Mar 6, 2009

gabbagabbahey

If c1=0, what can you say about c2? (Remember, the vectors are by assumption not zero vectors)

If c2=0, what can you say about c1?

The case where neither of them are zero was already covered in your 1st post.

6. Mar 7, 2009

bcjochim07

If c1=0, what can you say about c2? : c2 is nonzero

If c2=0, what can you say about c1? c1 is nonzero

so assume c1u + c2v = 0

if c1 is nonzero u = -c2/c1 * v
if c2 is nonzero v= -c1/c2 * u

part II

assume that u and v are scalar multiples of each other:

u = av and v = bu where both a and b are nonzero scalars and u and v are nonzero vectors

u - av = 0 and v - bu = 0 therefore u and v are linearly dependent

the problem also gives a hint that I should separately consider the case where one of the vectors is the zero vector

so assume u = av and v=bu

let u be the zero vector 1u-av = 0 therefore u and v are linearly dependent

let v be the zero vector 1v - bu= 0 therefore u and v are linearly dependent

I don't know if I'm getting this right at all.

7. Mar 7, 2009

HallsofIvy

Staff Emeritus
Yes, that is correct.

You are making the same mistake you did before. If u and v are scalar multiples of one another, then u= av and v= bu but it does NOT FOLLOW THAT "both a and b are nonzero scalars and u and v are nonzero vectors". If neither u nor v are zero, then you can say u= av for a non zero and so u- av= 0 with coefficients 1 and -a. If u= 0 then au+ 0v= 0 for any non-zero a and if v= 0, then 0u+ bv= 0 for any non-zero b.

8. Mar 7, 2009

de_brook

part i. If c1 is zero and c2 is not zero then for c1u + c2v = 0 to hold, either u = o(vector) or v = 0 and vice versa since it has been said that u and v are linearly dependent

9. Mar 7, 2009

de_brook

Is there a linear space V in which the union of any subspaces of V is a subspace except the trivial subspaces V and {0}? pls help