Show that a set of vectors spans a subspace of $\mathbb R^3$ even when it doesn't span $\mathbb R^3.$ Problem:
Show that $\{(1, 2, 3), (3, 4, 5), (4, 5, 6)\}$ does not span $\mathbb R^3.$ Show that it spans the subspace of $\mathbb R^3$ consisting of all vectors lying in the plane with the equation $x - 2y + z = 0.$
Attempt at the solution:
I made a matrix of:
$$A=\pmatrix{1&3&4\cr 2&4&5\cr 3&5&6\cr}$$ and reduced it to row-echelon form to determine rank. I found that rank$(A) = 2.$ Therefore, it can't span $\mathbb R^3.$
But then I need to show that it spans the plane given by the equation $x - 2y + z = 0.$
I thought that perhaps rewriting the problem as: $z = -x + 2y$ and plugging back in for $z.$ But that doesn't really tell me anything.
I also tried to show the span by demonstrating where it has linear combinations by writing span$(F)$ = span$\{ax - 2by + cz = 0\}$ but I don't know how to proceed.
Help?
 A: Row reducing (check details for yourself),
$$\pmatrix{1&3&4&|&x\cr 2&4&5&|&y\cr 3&5&6&|&z\cr}
  \sim\pmatrix{1&3&4&|&x\cr 0&-2&-3&|&y-2x\cr 0&0&0&|&x-2y+z\cr}\ .$$
Therefore
$$\eqalign{
  (x,y,z)\in\hbox{span(your vectors)}\quad
  &\Leftrightarrow\quad\hbox{the system has a solution}\cr
  &\Leftrightarrow\quad x-2y+z=0\ .\cr}$$
A: Make use of the fact that any plane $ax+by+cz=0$ is two dimensional. First note that all three vectors are in this plane. Hence they cannot span $\mathbb R^{3}$ (since the latter is 3 dimensional). The first two vectors are linearly independent by inspection because they are not scalar multiples of each other. Hence they span  a two dimensional subspace of the plane which implies that they span the entire plane. One way of showing that x-2y+z=0 gives a two dimensinal subspace is to note that $(1,0,-1)$ and $(2,1,0)$ are in it and they are independent, so the dimension of the plane is at least 2. The dimension cannot be 3 because the only subspace of dimension3 is $\mathbb R^{3}$ itself.
