I am doing a simple practice question:
Every combination of v = (1,-2,1) and w = (0,1,-1) has components that add to ____ . Find c and d so that cv + dw = (3,3,-6). Why is (3,3,6) impossible?
I know the blank is "zero" and c = 3, d = 9. The last part I didn't get --- (3,3,6) is impossible because it doesn't add up to zero.
What am I supposed to take away from this? Is there some property or definition here? (I'm starting linear algebra)
On
I would think of the problem this way: suppose you can move around $\mathbb{R}^3$ along the directions of the two vectors ${\bf v}=(1,-2,1)$ and ${\bf w}=(0,1,-1)$ (and let's say you start from the origin). Now, where can you go?
The point is that you can get from the origin to a point in this way if that point (or rather, the position vector of that point: the vector from the origin to the point) can be written in the form $c{\bf v}+d{\bf w}$ for some scalars $c$ and $d$ ("move $c$ units in the ${\bf v}$-direction and $d$ units in the ${\bf w}$-direction and you'll be there"). So this exercise is showing that you can get to $(3,3,-6)$ but you can't get to $(3,3,6)$.
Why this is a valuable question to ask, you'll learn later; for now, you're just exploring the way vectors let you move around the space (or, if you prefer, how vectors describe regions in the space) via linear combinations.
You have two linearly independent vectors $(1,-2,1)$ and $(0,1,-1)$ in the three dimensional space $\mathbb {R^3}$
Any linear combination of $(1,-2,1)$ and $(0,1,-1)$ is a vector whose components add up to $0$
The subspace spanned by these two vectors is a two dimensional subspace which does not include vectors whose components do not add up to $0$.
Therefore $(3,3,6)$ is not in that subspace since it is not a linear combination of $(1,-2,1)$ and $(0,1,-1)$