Vector Spaces & Subspaces

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 $W$.

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What have you tried? If this is homework, tag it as such. See meta.math.stackexchange.com/questions/1803/… –  lhf Oct 2 '12 at 23:44
Is it me or is asking $b$ completely meaningless if $W$ was not a subspace ? –  Belgi Oct 2 '12 at 23:57

(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.

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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 $y=x$.

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In order to prove (a) we need to show that:

1. $W\subseteq \mathbb{R}^3$
2. $(0,0,0)\in W$
3. Let $x,y\in W, \alpha \in \mathbb{R}$, implies $x+\alpha y \in W$.

(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 $w=(w_1,w_2,w_3)\in W\implies w=(w_1,w_2,w_2-w_1)=w_1(1,0,-1)+w_2(0,1,1)=sp\{(1,0,-1),(0,1,1)\}$

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I just have one question to The Q. Why is it w1- w2 and not the other way around? –  Sam Oct 3 '12 at 1:48
My bad, answer edited. Thank you! –  The-Q Oct 3 '12 at 8:20