Question

Let S be the subspace of R^3 consisting of all vectors of the form \[\vec v = (C_1, C_2, C_2-2C_1). Show that S is spanned by \(\vec v_1= (1, 0, -2)\) and \(\vec v_2 = (0,1,1)\)
Please, help

Answer 1

\[\vec{v}=(C_1,C_2,C_2-2C_1)\\
\vec{v_1}=(1,0,-2)\\
\vec{v_2}=(0,1,1)\]
Is this it?

Answer 2

yes.

Answer 3

Rewrite v as follows:
\[\vec{v}=(c_1,0,-2c_1)+(0,c_2,c_2)=c_1(1,0,-2)+c_2(0,1,1)\]
Thus, all vectors in S can be written as \(c_1\vec{v_1}+c_2\vec{v_2}\).

Answer 4

got you, much easier than my prof's lecture. thanks alot

Answer 5

No problem. :)

Answer 6

I am sorry for my bother, please help me convert: S = { (x - 2y -z =0: x, y , z in R}
find the set of vector spans S.

Answer 7

If you specify a value for y (say m) and z (say k), then you can see that x=2m+k so that every vector in S can be written as
\[\vec{u}=(2m+k,m,k)\]
You should know how to deal with this if you understood what I did with the other one. :)

Answer 8

@ybarrap

Answer 9

$$
\Large{
\vec{v_1}=(2,1,0)\\
\vec{v_2}=(1,0,1)\\
\vec u=m\times\vec{v_1}+k\times\vec{v_2}
}
$$

Answer 10

Does this make sense?

Answer 11

yes, but how to get? cannot just estimate, right? I know until z = -x +2y, and if set x = C_1, y = C_2, then the set become ???

Answer 12

Write the set with the smallest number of parameters, like we did with m and k above, then find two vectors with these as parameters. For example, we rewrote your vector as (2m+k,m,k). This only has two parameters, so we can know we can split this up into two independent vectors m(2,1,0) + k(1,0,1). There is no algorithm, but it;s like when you have a function f(x,y) and you parametrize like (x,y,f(x,y)), which now is a function of x,y only not (x,y,z). The number of variables you use to parametrize (like m,k or \(c_1,c_2\) will determine the number of basis vectors you need to span the space. In both cases we had two parameters so we needed to vectors to span the space.

Not sure if this made sense, but that's the basic idea.

Answer 13

It makes perfect sense. I got you.

Answer 14

I know it's basic idea but I have to have someone tells me the way to do, right? my prof use helicopter to lift me up to the information in the sky, therefore I don't know why I am there. hehehe

Answer 15

lol,ok

Answer 16

Split the vector so that each coefficient is by its lonesome. For u, it becomes u=(2m,m,0)+(k,0,k). You can see that m is alone, and so is k. You then factor out the m and the k.

Answer 17

thank you. I got it.