Linear Independence and Subspaces 
I'm currently looking at problem #2. I wrote $W$ in the form
$$W=a\begin{bmatrix}
2\\ 1\\ 4\end{bmatrix}+b\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}+c\begin{bmatrix}-1\\ 1\\ 
 1\end{bmatrix}$$
and then I took the determinant of the matrix formed by these vectors, that is,
$$\det\begin{bmatrix}2 & 1 &-1 \\  1&  0& 1\\  4&  1&1 \end{bmatrix}=0$$
Yet the study guide solutions say
$$ W=Col \begin{bmatrix}2 & 1 &-1 \\  1&  0& 1\\  4&  1&1 \end{bmatrix}$$
is a subspace of $\mathbb{R}^3$. I don't understand how it can be if these vectors are not linearly independent as the determinant is zero. What am I missing?
 A: The fact that they are not linearly independent does not mean that W is not a subspace. It is only mean that they are not the basis of the subspace - they aren't the minimal set that span w. . In order to show that W is subspace, you only need to take two vector, u and v in w, and show that u+v contains in w.
A: Okay so you got that $W$ is exactly equal to the solution set of this equation.
$$W=a\begin{bmatrix}
2\\ 1\\ 4\end{bmatrix}+b\begin{bmatrix}1\\ 0\\ 1\end{bmatrix}+c\begin{bmatrix}-1\\ 1\\ 
 1\end{bmatrix}$$
But this is saying that $W$ is equal to the span of those three vectors. Anytime you have the a subset defined by the span of a set of vectors, that subset is a subspace.
Theorem:
Let $W = span\{v_1,..,v_m\}$ be in $\mathbb{R}^n$. Then $W$ is a subspace of $\mathbb{R}^n$
Now, what YOU did was check to see if those vectors are linearly independent. Since $det=0$, they are not linearly independent. That just means that $W$ is not going to be a $3$ dimensional subspace (and so it isn't equal to the whole space, but rather it is a PROPER subspace).
A: You have already written
$W= a.x+b.y+c.z$, where $x=[2~ 1 ~4]^t$ $y= [1~ 0~ 1]^t$ and $z= c[-1~ 1~ 1]^t$, $a,b,c \in \mathbb{R}$.
That simply means $W$ is given by the column space of the mentioned matrix and that is what you guide's solution addresses. Note that column space of the matrix is a subspace of $\mathbb{R}^3$, since the matrix can be thought of a linear transformation from $\mathbb{R}^3$ to $\mathbb{R}^3$ and colum space of the matrix is the range of the linear transformation.
Note that "determinant of matrix is zero" means that the vectors $x$, $y$, $z$ are not linearly independent. However, it does not deny the fact that $W$ is spanned by them. Of course, the set of vectors $\{x,y,z\}$ is not the minimal spanning set of $W$, i.e., not a basis of $W$. If it was, you would have obtained the determinant to be non-zero.
