Find a basis and the dimension of the solution space $\textsf{W}$ $$\left\{\begin{align}
     x + 3y + 2z = 0 \\
     x + 5y + z  = 0 \\
     3x + 5y + 8z = 0 \\
\end{align}\right.$$
So if we represent this as an augmented matrix
$$\begin{pmatrix}
     1 & 3 & 2 & 0 \\
     1 & 5 & 1 & 0 \\
     3 & 5 & 8 & 0 \\
\end{pmatrix}$$
In row reduced form would be
$$\begin{pmatrix}
     1 & 0 & \tfrac{7}{2} & 0 \\
     0 & 1 & -\tfrac{1}{2} & 0 \\
     0 & 0 & 0 & 0 \\
\end{pmatrix}$$
Therefore, a basis of the set we can say would be vector
$$\begin{pmatrix} 1 \\ 1 \\ 3 \end{pmatrix}$$
and the vector $$\begin{pmatrix} 3 \\ 5 \\ 5 \end{pmatrix}$$
right? I'm new to these problems and want to make sure I've got the right idea in approaching the solution.
 A: It is not correct. How did you obtain these basis vectors?
By row reduction you reduced your system to
$$\begin{cases} x_1 + \frac72x_3 = 0 \\ x_2 - \frac12x_3 = 0\end{cases}$$
so $$\begin{bmatrix} x_1 \\ x_2 \\ x_3\end{bmatrix} = t\begin{bmatrix} -\frac72 \\ \frac12 \\ 1\end{bmatrix}, \quad\text{ for some } t \in \mathbb{R}$$
so e.g. $\left\{\begin{bmatrix} -7 \\ 1 \\ 2\end{bmatrix}\right\}$ is a basis for the solution space.
A: If you subtract the second equation from the first you have that 
$x-x+5y-3y+z-2z=0$
So
$z=2y$ 
Now if you change $z$ with $2y$ in the third equation you have that 
$3x+5y+16y=0$
$x=-7y$
Now if you change $x$ with $-7y$ in the first equation you have that 
$-7y+3y+4y=0$ for all $y\in \mathbb{R}$ 
So the set of solution of your system will be 
$W=\{\lambda (-7, 1,2 ): \lambda \in \mathbb{R}\}$ 
So a base of your subspace will be $\{(-7,1,2)\}$
A: If you check your work by plugging the coordinates of the two vectors that you’ve found back into the original system, you’ll see that neither one is a solution. It looks like you’ve gotten the column space of a matrix $A$ confused with its null space. The latter is what you’re looking for: it’s the set of vectors that are solutions to $A\mathbf x=0$, which is what you’re looking for. The column space, on the other hand, is the subspace spanned by the columns of $A$. This happens to be the set of vectors $\mathbf b$ for which $A\mathbf x=\mathbf b$ has a solution—in other words, the set of right-hand sides for which the system of linear equations is consistent.  
It’s much harder to get these two mixed up when the coefficient matrix isn’t square. Observe that your row-reduction showed that the last equation is a combination of the first two, so we can remove it entirely from the system right away without affecting the solution (that’s effectively what the row-reduction did). If we did that, we’d be starting with the augmented matrix $$\left[\begin{array}{ccc|c}1&3&2&0\\1&5&1&0\end{array}\right]$$ instead. This reduces to the same matrix that you found, but without the last zero row. Taking the pivot columns as before, we get the vectors $(1,1)^T$ and $(3,5)^T$, neither of which could possibly be a solution to the system since we need elements of $\mathbb R^3$, not of $\mathbb R^2$.  
To find a basis for the null space of the matrix, you need to examine the non-pivot columns of the reduced matrix. That’s the third column, from which you can read the basis vector $(7/2,-1/2,-1)^T$. If you substitute these coordinates into the system of equations, you’ll see that this is indeed a solution to it.  
Note, by the way, that once you’ve gotten used to doing these computations there’s no particularly good reason to augment the coefficient matrix when you’re working with a homogeneous system. That last column will remain zero no matter what elementary row operations you perform.
