Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $W$.

share|improve this question
2  
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

3 Answers 3

(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.

share|improve this answer

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 $y=x$.

share|improve this answer

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

share|improve this answer
    
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.