How to test if these are a vector space and find the basis? I have been trying to work through these linear algebra questions in my text book for hours now, but i just cant seem to figure it out.
The question is:

Is X or Y a vector space? If so, what is its basis?

I cant seem to figure out how to prove/show this no matter what i do.
If someone wouldnt mind giving me some pointers i would be hugely grateful.
Thanks
Corey
 A: Well, I'd rather try to guess the right answer from geometrical point of view before trying to solve it analytically.
In your example, Y is a subspace of vectors such that their projection on $v = (1 ; 4 ; 3)$ is zero, that means that they are orthogonal to $v \Rightarrow$ that is a 2D-plane in 3D space. If we pick few random points from a 2D-plane in 3d-space and let's say, try to find their average, would it still be on a plane - sure it would, that means that space of points on that plane is invariant wrt averaging, which is good and make us assume that this space is likely to be vector linear space. The same thing  applies to vector product ($\times$), as soon as the length of the vector you get after vector product is equal to the measure of the parallelogram they bound (=0 in your case) $\Rightarrow$ they much be collinear, that seem to be a good subspace too.
Using such geometrical interpretation you could guess the right answer and after that try to prove it just like A.P. or Nizar suggested.
A: The set $X$ is the set of vectors linear dependent to the vector $$\pmatrix{1\\4\\3}$$
So, $X$ is a vectorspace. The given vector is a basis.
$Y$ is a plane containing the origin. So, it can be written
as $E(x)=ru+sv$, with linear independent vectors $u$ and $v$.
So, $Y$ also is a vector space. For example, the vectors $\pmatrix{-4\\1\\0}$ and $\pmatrix{-3\\0\\1}$ form a basis.
A: Hint: Both the cross product and the dot product by a fixed vector are linear operations, i.e.
$$
\begin{gather}
(a\mathbf{x} + b\mathbf{y}) \times \mathbf{v} = (a\mathbf{x}) \times \mathbf{v} + (b\mathbf{y}) \times \mathbf{v} = a(\mathbf{x} \times \mathbf{v}) + b(\mathbf{y} \times \mathbf{v})\\
(a\mathbf{x} + b\mathbf{y}) \cdot \mathbf{v} = (a\mathbf{x}) \cdot \mathbf{v} + (b\mathbf{y}) \cdot \mathbf{v} = a(\mathbf{x} \cdot \mathbf{v}) + b(\mathbf{y} \cdot \mathbf{v})
\end{gather}
$$
for every $\mathbf{x},\mathbf{y} \in \Bbb{R}^n$ and for every $a,b \in \Bbb{R}$, with $\mathbf{v} \in \Bbb{R}^n$ fixed.
A: Let us start with $Y$. we have 


*

*$ [0,0,0] \begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix}=0 $ hence the vector  $0_{ \mathbb{R^3} }  \in Y$.

*let $u,v \in Y$, then we have  $ (u+v)\begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix}= u\begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix}+ v\begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix} = 0+0=0 $. Hence  $u+v \in Y$.

*Let $\alpha \in  \mathbb{R} $ and  $u \in Y$, then we have $ (\alpha u) \begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix}= \alpha \Bigg( u \begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix} \Bigg)= \alpha (0)=0 $. Thus $\alpha  u \in Y$.
Therefore $Y$ is a subspace of $\mathbb{R^3} $. Thus $Y$ is a vector space .
To find its basis, we may do the following :
For any $[a,b,c] \in Y$ we have $$ [a,b,c] \begin{bmatrix} 1 \\ 4 \\ 3  \end{bmatrix}=0$$This implies $$  a+4b+3c=0  \; \Rightarrow \; a= -4b-3c
 $$
Thus every vector $u$ in $Y$ is 
$$u=[a,b,c]= [-4b-3c,b,c]= b[-4,1,0]+ c[-3,0,1]  $$
Thus  $\{ [-4,1,0], [-3,0,1] \}$ is a system of generator of $Y$, however these two vectors are also linearly independent, hence they form a basis for  $Y$. 
Similarly you may proceed for  $X$. 
