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Let S be the collection of vectors in $[x,y,z]$ in $R^3$ that satisfy the given property. In each case, either prove that $S$ forms a subspace of $R^3$ or give a counter example to show that it does not. Case: $z = 2x, \, y=0$

Okay, there are 3 conditions that need to be satisfied for this to work.

Zero vector has to be a possibility: Okay, we can find out that this is true. $[0,0,0]$ E S

Addition between two vectors: $[x_1,0,2x_1] +[x_2,0,2x_2] = [x1+x2,0,2(x1+x2)]$,Yes that works

Scalar multiplication: $c[x,0,2x] = [cx,0,2cx]$, okay that works as well..

So, yeah it should be a subspace in $R^3$.

But my doubts are when I graph this onto wolfram alpha, I see that it gives me a 2D graph. Do you guys see an error in my logic above? I think I am right,

Thanks

Muphrid
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A A
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  • I don't see an error, it is indeed a subspace. What exactly is wrong with the plot of this subspace?... – Ludolila Mar 18 '14 at 18:02
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    Yes you are right. This is definitely a subspace. You are also right in saying that the subspace forms a plane and not a three-dimensional locus such as $\Bbb R^3$. But that should not be a problem. As long as this is a set which satisfies the axioms of a vector space we are fine. Arguments are fine. Answer is correct in my opinion. – Ishfaaq Mar 18 '14 at 18:02
  • I just seen a 2D graph when plotted, and I became worried and I didn't want to get this wrong. Thanks guys, very much appreciated. – A A Mar 18 '14 at 18:03
  • The space is a line. The line lies in the $xz$-plane since you are constrained to $y=0$. It is the familiar graph of the line $z=2x$, regarded as lying in $\mathbb R^3$. – MPW Mar 18 '14 at 18:04
  • @MPW: Right you are.. My Bad.. – Ishfaaq Mar 18 '14 at 18:08

1 Answers1

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Notice that (${\bf v} \equiv (1,0,2)$): $$S=\{t{\bf v}:t \in \mathbb{R}\}= span\{{\bf v}\}$$ And it is easily shown (And known) that a linear span fulfills all of the vector space axioms.

Alternatively, there is the succinct subspace criterion:

If $V$ is a vector space over a field $F$, and $W \subseteq V$, then $W$ is a subspace of $V$ if and only if these $2$ conditions hold:

  1. $W \neq \emptyset$
  2. For all $a,b \in F$ and ${\bf x},{\bf y} \in W$: $a{\bf x} + b{\bf y} \in W $

In our case:

  1. ${\bf 0}=0{\bf v} \in S \implies S \neq \emptyset$
  2. For all $a,b \in \mathbb{R}$ and ${\bf x},{\bf y} \in S$, there exist $t_1,t_2 \in \mathbb{R}$ such that ${\bf x}=t_1{\bf v},{\bf y}=t_2{\bf v}$ and therefore $a{\bf x} + b{\bf y}=(at_1+bt_2){\bf v} \in S $, since $at_1+bt_2 \in \mathbb{R}$.
user76568
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