Question

For the vectors v1=(1,0,2,2) v2=(1,1,0,3), v3=(1,4,0,0) find a linear homogeneous system which has the span of these vectors as solution space.

Answer 1

@ivanmlerner
do you know how to do this one?

Answer 2

Just a second, I need to look some words in the dictionary, I'm not familiar with this subject in english.

Answer 3

Look, I have a way, but since I haven't had linear algebra beyond the most basics this is probably not the best approach.
We know that the solutions of a homogeneous linear system are given by av1+bv2+cv3
And that this linear system must have the following form:
a1x+b1y+c1z+d1h=0
a2x+b2y+c2z+d2h=0
a3x+b3y+c3z+d3h=0
a4x+b4y+c4z+d4h=0

Lets take the first of these equations, now we have 3 points that must satisfy this equation and that are linearly independent(not sure if that is the term you use), and we can get a fourth just by making a linear combination of these 3.
Doing this, we get an equation with 4 unknown constants and four points that mus satisfy this equation, therefore we have a 4x4 system with a1, b1, c1, d1 as the solution.

Doing that for all of the four equations, we find all the constants needed.

Answer 4

so are you saying that the system we are finding is one equation, or 4 equations?

Answer 5

4 equations to find the 4 constants of the first equation of the original system, and the same thing for the other 3

Answer 6

i was thinking that you set those vectors in a matrix and you do reduced echelon form, but didnt know what to do after that...

i am not sure if i quite understand how to make the four equations ....
what are the four equations i would use for (1,0,2,2)

Answer 7

I don't know what this method that you said is, because as I said I haven

Answer 8

what course did you take that you learned some of this stuff from?

Answer 9

t studied that so far, but what I was thinking was not make 4 equations for each poin, but make for equations using all points for each equation of the system

Answer 10

Highschool, just some simple methods for solving linear systems and nothing greater than 3 dimensions, but even though I don't have any of the other methods, mine usually work, even though sometimes they are not very eficient

Answer 11

well if they work, then id like to learn how to use them, but i don't think I quite understand how to do your method.

Answer 12

the system you want is this:
a1x+b1y+c1z+d1h=0
a2x+b2y+c2z+d2h=0
a3x+b3y+c3z+d3h=0
a4x+b4y+c4z+d4h=0

Looking just at the first equation and suposing all points and its linear combinations are solutions of it:
a1+2c1+2d1=0 (point (1, 0, 2, 2))
a1+b1+3d1=0 (point (1, 1, 0, 3))
a1+4b1=0 (point (1, 4, 0, 0))
2a1+b1+2c1+5d1=0 (Point v1+v2=(2, 1, 2, 5))
Since the span of these vectors must be a solution v1+v2=(2, 1, 2, 5) must also be a solution.
Now you have a linear system wich solution gives you a1, b1, c1, d1.
This system so not have only one solution, but thats ok, since you want a system that gives you some restrictions so that you can create a system that has the span os these points as a solution. To sumarize, your problem has many valid solutions.

Answer 13

is there a reason we have to have 4 equations for our system? why cant we have just one?

Answer 14

Well, we have 3 vectors that are independent, and all of them must satisfy the equation so I was wrong to "create" a point to make a fouth equation, but you need to have at least 3, because otherwise you are not aplying all the conditions and the equation might not satisfy all the points, giving a wrong answer.

Answer 15

If I gave you a vector, lets call it "u"... and if i was to multiply this vector by v1, v2, and v3 such that v1u=v2u=v3u=0..... would that satisfy what I am looking for or no??

Answer 16

You mean a vector that is perpendicular to all the three? I thought about it too because the equation is similar to the equation of a plane, but I got consused because of the 4 dimensions, so I just followed another way, but maybe that is true.

Answer 17

yeah im not really sure either... i do thank you for your help though. it sounds like you understand this stuff better than i do, and im in the class.

Answer 18

you are welcome, the methods you are learning are much more powerfull though, let me know if you get the result with this method