Let $W = \{ (x,y,z) \in \mathbb{R}^3 : x-y+z=0 \}$.
a) Is $W$ a subspace of $\mathbb{R}^3$?
b) Find a spanning set for $W$. Give a complete geometric description of $W$.
Let $W = \{ (x,y,z) \in \mathbb{R}^3 : x-y+z=0 \}$.
a) Is $W$ a subspace of $\mathbb{R}^3$?
b) Find a spanning set for $W$. Give a complete geometric description of $W$.
(a) Yes.
(b) It is the plane consisting of vectors whose inner products with $(1,-1,1)$ are zero. Take any two such vectors so that the two are not parallel.
To elaborate on Tom's answer, it helps to take this one backwards. If you look at the geometry first, the other answers fall into place. The mathematical definition of a vector in $\mathbb R^n$ is essentially the same as a point in $\mathbb R^n$, so we can imagine that the vectors in $W$ are points that fit the condition that $x-y+z=0$ or $z=y-x$, which is clearly a plane passing through the origin. Any two non-collinear vectors in that plane will span the plane. So that takes care of (b).
For (a), we just need to establish that the zero vector is a member of $W$, and that it is the same zero vector as in $\mathbb R^3$, and that $W$ is closed under vector addition and scalar multiplication. A plane passing through the origin contains the zero vector, of course. And the ease of finding a spanning set shows us that the set is closed under the same vector operations in place for $\mathbb R^3$. To prove it you would just have to add two arbitrary vectors in $W$ and show that the components obey $x-y+z=0$.
Also, as Belgi points out, asking (b) in the first place gives away that $W$ is a subspace.
Edit: You know, if you have to "completely" describe the geometry, it's probably worth noting that the plane $W$ describes is tilted $45^{\circ}$ above the $x - y$ plane and crosses the $x - y$ plane along the line $y=x$.
In order to prove (a) we need to show that:
(1.) Trivial.
(2.) Trivial.
(3.) $x+\alpha y = (x_1,x_2,x_3)+\alpha(y_1,y_2,y_3)=(x_1+\alpha y_1, x_2+\alpha y_2, x_3+\alpha y_3)\implies$ $\;\;\;\;x_1+\alpha y_1-(x_2+\alpha y_2)+(x_3+\alpha y_3)=0$ (since $\;\;x,y \in W$) $\implies x+\alpha y \in W$
(1)+(2)+(3)$\implies W$ is a subspace of $\mathbb{R}^3$.
In order to prove (b): $x-y+z=0$ so for every $w=(w_1,w_2,w_3)\in W\implies w=(w_1,w_2,w_2-w_1)=w_1(1,0,-1)+w_2(0,1,1)=sp\{(1,0,-1),(0,1,1)\}$