# Homework Help: Linear Algebra vector functions LI or LD

1. Jul 20, 2015

### Fellowroot

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

Determine whether or not the vector functions are linearly dependent?

u=(2t-1,-t) , v= (-t+1,2t) and they are written as columns matrixes.

2. Relevant equations
Wronskian, but I don't think I should use it because I need to take derivatives so it doesnt seem like it would work.

3. The attempt at a solution

I took the determinate and got (2t-1)(2t) - (-t)(-t+1) =0

got t=0 and t= 1/3

but I dont really know if im even doing this correctly. I know that if we have just values instead of functions then if the determinate equals 0 then that means that its linear dependent, but again in this case I'm clueless.

Here are two values which make the det zero but not all values of t make the det zero. Any help is appreciated.

2. Jul 21, 2015

### Staff: Mentor

Since the two vector-valued functions are not scalar multiples of each other, the two are obviously linearly independent. Regarding your approach, for the two functions to be linearly dependent, the determinant would have to be identically zero, not just zero for a couple of values of the parameter.

3. Jul 21, 2015

### Fellowroot

Thank you for the response, but this only leads me to another question.

So the next question is this:

u= (t-t^2, -t) v=(-2t+4t^2,2t) and it asks the same thing as stated in my original post.

So clearly one is not a multiple of another, yet when I find det I get -6t^3=0 so the only solution is that t must be zero for all t, but in the back of the book it says that its actually linear independent?

So does this mean that all I really have to do is look at the two vectors and check if one is a scalar multiple of the other????

4. Jul 21, 2015

### HallsofIvy

On the contrary, the second clearly is -2 times the first!

That is, after all what linearly independent means!

Don't become so focused on "formulas" that you forget the basic definitions. A set of vectors, $\{v_1, v_2, ..., v_n\}$, is "independent" if and only if the only set of numbers, $\{A_1, A_2, ..., A_n\}$ such that $A_1v_1+ A_2v_2+ ...+ A_nv_n= 0$ is $A_1= A_2= ...= A_n= 0$. If not, the set is dependent.
In particular, for two vectors, $A_1v_1+ A_2v_2= 0$ with $A_2\ne 0$ gives $v_2= \frac{A_1}{A_2}v_1$ so, yes, two vectors are dependent if and only if one is a multiple of the other.

5. Jul 21, 2015

### Fellowroot

actually i dont think that is so.

t-t^2 is not a -2 multiple of -2t+4t^2

the 4 makes it not work like this.

Also on number 5 in my homework I have this problem:

u = ( 2-t,t,-2) v = (t, -1, 2 ) w = (2+t, t-2, 2 )

it says in the answer is linear dependent, yet how can this be? I do not see how one vector is a scalar multiple of another?

can someone explain? Thanks.

Last edited: Jul 21, 2015
6. Jul 21, 2015

### DeldotB

Can you combine two of the three to create the third? (Like BIG HINT: -2v+w =u)
So {u,v,w} is L.D

Also, I computed the determinant and the set is identically zero for all t thus L.D

Last edited: Jul 21, 2015
7. Jul 21, 2015

### Fellowroot

Okay I'm still trying to master this topic. Can someone please explain to me how they got linear dependent on this following problem.

u = (e^t,0)
v = (0,0)
w = (0, e^t)

How is this dependent? I cannot for the life of me find something like au + bv = w.

8. Jul 21, 2015

### DeldotB

What do you know about sets that contain the zero vector? Are they always L.D or L.I?
Think about the defiition of Linear Independence. Can a set conatining the zero vector ever be L.I?

9. Jul 21, 2015

### DeldotB

Say {v1, v2, . . . , vn} is a finite set of vectors and one of them is the zero vector.
So just let v1=0.
It we let constnats c1 = 1 and c2 = · · · = cn = 0,
this imples c1v1 + c2v2 + · · · + cn + vn = 1 · 0 + 0 · v2 + · · · + 0 · vn = 0
So what does this mean?

10. Jul 21, 2015

### Staff: Mentor

For linear dependence, the equation au + bv + cw = 0 must have multiple, nonzero solutions for the constants a, b, and c. It should be obvious that that is the case here.