MHB Spanning Spaces with Linear Combinations

[IrinaK.]

Hi Christina! Welcome to MHB! :)

If all vectors are a multiple of each other, they form a line through the origin.

If 2 vectors are independent, that is, not a multiple of each other, they "span" a plane.

If 3 vectors are independent, that is, the 3rd can not be written as the sum of multiples of the other 2 vectors, they "span" all of R3.
Note that if the 3rd vector can be written as the sum of multiples of the other 2 vectors, that vector effectively belongs to the plane those 2 vectors "span".
We say that the 3rd vector is a "linear combination" of the other 2.

Hello!
Christina, thank you for this thread. I also have the same issue.

I like Serena, I would be grateful for your further help.
If given vectors are (1,2,3) and (3,6,9) that have v3 = v1+v2, does it mean that the whole V vector or W vector lies only on XOY plane (two dimensional) and OZ is not applicable, or this is an incorrect assumption? I have just started linear algebra and lack some basic understanding.
Thank you!

 

[Siron]

Hi,

To solve your problem it would be nice if you could explain the variables you're using. What is v1? v2? v3? V vector? W vector?

 

[IrinaK.]

Hi,

To solve your problem it would be nice if you could explain the variables you're using. What is v1? v2? v3? V vector? W vector?

Siron,

I assume that v1, v2, v3 are for vector V (first one in my example) and vector W is the second one. I didn't think that it is important to state that :)

 

[Fallen Angel]

Hi,

This is my guess of what you want to ask.

You got, two vectors, namely $v=(1,2,3)$ and $w=(3,6,9)$.

They both have $v_{3}=v_{2}+v_{1}$, or using common notation, $z=x+y$.

And you ask if they lie in the plane $XOY$, I understand that this is the plane formed by the $X$ and $Y$ axis, this is $XOY=\{(x,y,z)\in \Bbb{R}^{3} \ : \ z=0\}$.

The answer to this is NO, for being in this plane they should have $z=0$.

But they both lie in the same plane, the plane given by the first equation, which is $\pi =\{(x,y,z)\in \Bbb{R}^{3} \ : \ x+y=z \}$ (Remember that the dimension of a subspace is the dimension of the ambient space minus the number of linearly independent equations used to define the subspace, which in this case is 1).

Furthermore, this doesn't implies that $v,w$ define a plane, actually $w=3v$, it is, they are linearly dependent, so they both define a line. $\langle v,w\rangle=\langle v\rangle $