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I marked my answers but I am incorrect.

Here's my logic -

A. $0$ is not a solution for this equation so not a subspace.

B. ... not sure about this one - Is this a subspace of $\mathbb{R}^3$?

C. Closed under multiplication and addition so it is a subspace.

D. If I write this as $V = (-y-z,y,z)$ I can see this set of vectors is closed under multiplication and addition so it is a subspace.

E. If I let $x = 1$ I have $ V_1 =(1, -1, -8)$, If I let $x = 2$ I have $V_2 =(2, 0, -9)$

The sum of these vectors is $(3, -1, -15)$ which is not an element of the original set so this is not a subspace.

So where am I going wrong?

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1  
Take a look at the answer to your previous question: math.stackexchange.com/questions/122033/… –  Shaun Ault Mar 19 '12 at 11:46
    
Cheers for the explanation on the other question mate. –  Jim_CS Mar 19 '12 at 12:05

3 Answers 3

up vote 1 down vote accepted

You're just about there.

A) Correct. Checking whether the potential subspace contains the zero vector is a great place to start on these sorts of problems. No zero vector -- not a subspace.

B) This is a subspace. The set of solutions of a homogeneous linear system is always a subspace (it is known as a "null space" or "kernel" of some corresponding linear transformation). In fact, if you let $M = \begin{bmatrix} 2 & 9 & 0 \\ 8 & 0 & -5 \end{bmatrix}$, then the set in B) is just $\mathrm{Null}(M)$ (the nullspace of $M$). Again, Nullspaces are always subspaces.

Alternatively, you could solve the corresponding system $M{\bf x}={\bf 0}$ and find that elements of that set look like ${\bf x} = \begin{bmatrix} 5/8 \\ -5/36 \\ 1 \end{bmatrix}z$ for any choice of $z \in \mathbb{R}$. So this set is the span of $\begin{bmatrix} 5/8 \\ -5/36 \\ 1 \end{bmatrix}$ (and thus a subspace since all spans are subspaces).

C) Correct. You can see this is a subspace because the elements are just multiples of $(-5,3,4)$ and again spans are subspaces.

D) Correct. This is a subspace for the same reasons as part B) (it's the set of solutions of a homogeneous linear system).

E) Correct. Not a subspace. Easier counterexample: $(1,1,1)$ is in there. But $(-1)(1,1,1)=(-1,-1,-1)$ is not (so not closed under scalar multiplication).

F) Not a subspace. $(0,0,0)$ is not in there.

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For $(B)$, you could also note that both defining equations are planes through the origin, hence their intersection is either a line or plane through the origin. (The intersection is a line in this case since the normal vectors are different.) Therefore the intersection is a subspace.

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HINT: Any of a subspace of $\mathbb{R}^3$ is belong to one of a following.
1) $\{(0,0,0)\}$
2) Line through the origin.
3) Plane through the origin.

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