What is the solution to finding spanned subspaces?

[Telemachus]

Hi, I'm trying to solve this exercise.

Homework Statement

Given the following subsets of $$\mathbb(R)^3$$ find the subspaces generated by them:
$$\{(1,0,-2),(-1,0,2),(3,0,-1),(-1,0,3)\}$$

The Attempt at a Solution

I've tried to solve the linear dependence, so I've made the system:
$$\begin{Bmatrix} a-b+3c-d=0 \\-2a+2b-c+3d=0 \end{matrix}$$

And I got: $$\begin{Bmatrix} a-b+8c=0 \\5c+4d=0 \end{matrix}$$

$$a=b-8c$$ and $$d=-5c$$

$$(b-8c,b,c,-5c)$$

I don't know what to do next. I know that there is a linear dependence, but I don't know how to work with it, and what to do with the result I've found. Does it mean that the solution is a plane because there are only two free variables in $$(b-8c,b,c,-5c)$$? and how do I get the solution from here.

Bye there, and thanks of course.

 

[Mark44]

Hi, I'm trying to solve this exercise.

Homework Statement

Given the following subsets of [\ mathbb (R) ^ 3] find the subspaces generated by them:
$$\{(1,0,-2),(-1,0,2),(3,0,-1),(-1,0,3)\}$$

The Attempt at a Solution

I've tried to solve the linear dependence, so I've made the system:
$$\begin{Bmatrix} a-b+3c-d=0 \\-2a+2b-c+3d=0 \end{matrix}$$

And I got: $$\begin{Bmatrix} a-b+3c-d=0 \\-2a+2b-c+3d=0 \end{matrix}$$ and $$d=-5c$$

Where did the d = -5c come from? How did you go from two equations to the same two equations plus one more?

$$(b-8c,b,c,-5c)$$

I don't know what to do next. I know that there is a linear dependence, but I don't know how to work with it, and what to do with the result I've found. Does it mean that the solution is a plane because there are only two free variables in $$(b-8c,b,c,-5c)$$? and how do I get the solution from here.

Bye there, and thanks of course.

 

[Telemachus]

Lets see:

$$\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{-2}&{2}&{-1}&{3}\end{bmatrix}$$
Twice the first row plus the second on the second:
$$\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{0}&{0}&{5}&{1}\end{bmatrix}$$

$$\begin{bmatrix}{1}&{-1}&{8}&{0}\\{0}&{0}&{5}&{1}\end{bmatrix}$$

Thats what I did.

Srry, there was something wrong in my first post. I've just corrected it.

 

[Mark44]

Lets see:

$$\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{-2}&{2}&{-1}&{3}\end{bmatrix}$$
Twice the first row plus the second on the second:
$$\begin{bmatrix}{1}&{-1}&{3}&{-1}\\{0}&{0}&{5}&{1}\end{bmatrix}$$

The matrix above is fine.

$$\begin{bmatrix}{1}&{-1}&{8}&{0}\\{0}&{0}&{5}&{1}\end{bmatrix}$$

This didn't do you any good. You want to use leading entries in a row to eliminate nonleading entries in the rows above and below. Multiply the 2nd row to get 0 0 1 1/5, and use the leading entry in the second row to eliminate the 3 in the row above it. That will give you a completely reduced, row-echelon matrix.

Thats what I did.

Srry, there was something wrong in my first post. I've just corrected it.

The last step you did wasn't technically wrong - it just wasn't a big help.

Since the set of vectors you started with had four vectors in R³, there are obviously too many to form a basis for R³. By inspection you can see that the second vector is a multiple of the first.

It's not so easy to see (the work you are doing shows it), but the 4th vector is a linear combination of the first three. This means that the subspace spanned by the four vectors is the same as that spanned by the first and third.

 

[Telemachus]

Thank you very much Mark.

So, what I got is:

$$\begin{bmatrix}{1}&{-1}&{0}&{-8/5}\\{0}&{0}&{1}&{1/5}\end{bmatrix}$$

How should I use this?

 

[HallsofIvy]

I prefer your original form:
$$\a(1,0,-2)+ b(-1,0,2)+ c(3,0,-1)+ d(-1,0,3)$$
Although I would NOT use that to "solve the independence" (you KNOW they are not independent because there are 4 vectors in $R^3$).

Any vector in that subspace can be written
$$(x, y, z)= a(1,0,-2)+ b(-1,0,2)+ c(3,0,-1)+ d(-1,0,3)$$
or
$$(x, y, z)= (a- b+ 3c- d, 0, -2a- 2b- c+ 3d)$$

One obvious result is that y= 0. Now, what further relationships can you find between x and z?

 

[Telemachus]

Well, $$x=a-b+3c-d$$ and $$z=-2a-2b-c+3d$$

And as I said $$a=b-8c$$ and $$d=-5c$$, so...

$$x=b-8c-b+3c-5c=-10c$$ and $$z=-2(b-8c)-2b-c+3(-5c)=-2b-16c-2b-c-15c=-32c-4b$$