Linearly Dependent: Show {1,2,1,0},{3,-4,5,6},{2,-1,3,3},(-2,6,-4,-6)

[jesuslovesu]

Homework Statement

Show that the given set is linearly dependent and write one of the vectors as a linear combination of the remaining vectors.

{(1,2,1,0), (3,-4,5,6), (2,-1,3,3), (-2,6,-4,-6) }

Homework Equations

The Attempt at a Solution

I've tried setting up equations like
(0,0,0,0) = c1 * (1,2,1,0) + c2 * (3,-4,5,6) + c3 * (2,-1,3,3) + c4 * (-2,6,-4,-6)
0 = c1 + 3c2 + 2c3 - 2c4 ... etc
Unfortunately each time I try to solve them, I keep on getting 0 = 0.

But I've noticed that if I set the constant (c3) in front of the third vector (v3) in the set to 0 then it works.
v4-v1 + v2 = (0,0,0,0). Is that a valid way to answer this question? Can I set one of the constants to 0?

And if so, why can't I just set all the constants to 0 and be done with it?

 

[learningphysics]

Did you set it up as an augmented matrix and use gaussian elimination? The fact that you get a row with all 0's shows that the vectors are linearly dependent...

If the only solution is c1=c2=c3=c4=0 (which is always a solution whether the vectors are dependent or independent), then the vectors are linearly independent. so showing that setting all constants to zero solves the system doesn't help to show they are dependent...

If you have a row with all 0's, then you have a free variable (if this sounds unfamiliar check your text about this)... choose c4 = 1... now get the other values... then write the fourth vector in terms of the first 3.

 

[HallsofIvy]

Homework Statement

Show that the given set is linearly dependent and write one of the vectors as a linear combination of the remaining vectors.

{(1,2,1,0), (3,-4,5,6), (2,-1,3,3), (-2,6,-4,-6) }

Homework Equations

The Attempt at a Solution

I've tried setting up equations like
(0,0,0,0) = c1 * (1,2,1,0) + c2 * (3,-4,5,6) + c3 * (2,-1,3,3) + c4 * (-2,6,-4,-6)
0 = c1 + 3c2 + 2c3 - 2c4 ... etc
Unfortunately each time I try to solve them, I keep on getting 0 = 0.

But I've noticed that if I set the constant (c3) in front of the third vector (v3) in the set to 0 then it works.
v4-v1 + v2 = (0,0,0,0). Is that a valid way to answer this question? Can I set one of the constants to 0?

And if so, why can't I just set all the constants to 0 and be done with it?

I don't understand what you mean by "getting 0= 0". Exactly what are you doing to solve them? You can't just "set all the constants to 0 and be done with it" because the whole point is whether there is ANOTHER way of writing the 0 vector. If a set of vectors "spans" the space, then every vector can be written as a combination of those vectors. If a set of vectors is DEPENDENT then some vectors can be written as a combination of those vectors in more than one way. If a set of vectors both spans the space and is independent, then every vector can be written as a combination of them in ONLY ONE WAY

You say "I've noticed that if I set the constant (c3) in front of the third vector (v3) in the set to 0 then it works." I'm not sure what you mean by "works". Setting c3 to 0 in (0,0,0,0) = c1 * (1,2,1,0) + c2 * (3,-4,5,6) + c3 * (2,-1,3,3) + c4 * (-2,6,-4,-6) gives (0,0,0,0)= (0,0,0,0) = c1 * (1,2,1,0) + c2 * (3,-4,5,6) + c4 * (-2,6,-4,-6) .
Then what? you have 0= c1+ 3c2-2c4, 0= 2c1-4c2+6c4, 0= c1+ 5c2-4c3, 0= 6c2-6c4.
That last equation, of course, says c2= c4 so the other equations become 0= c1- c2, 0= 2c1+ 2c2, and 0= c1+ c2. Adding the first and last of those, 0= 2c1 so c1= 0. and then c2= c4= 0. Okay, assuming c3=0, you get that all the other cooefficients 0 also. But what if c3 is not 0? It is still possible that there would be some combination with c3 not 0 that would work.