# Do these vectors form a basis for $x_1 + 2x_2 + 3x_3 + 6x_4=0$?

Are the vectors

$$\begin{bmatrix} -2\\ 1 \\ 0 \\ 0\end{bmatrix}, \begin{bmatrix} -1\\ -1 \\ 1 \\ 0\end{bmatrix}, \begin{bmatrix} -4\\ -2 \\ 2 \\ 1\end{bmatrix}$$

a basis for the subspace described by the 3-dimensional hyperplane $x_1 + 2x_2 + 3x_3 + 6x_4=0$?

My understanding of a basis isn't very good. I initially thought that the vectors could not form a basis because there are only 3 of them, and to fully span a vector space (one of the requirements of a basis), there needs to be as many vectors as there are dimensions, and the dimension is 4 because that is the number of rows in each vector. Then the formula for the hyperplane (not sure what the significance of this is either) is "3 dimensional". I'm kind of lost on this one and I don't have the solution for it to help me understand.

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The dimension is certainly not four. You can pick $x_1,x_2,x_3$ arbitrarily (that's three degrees of freedom), but that fully determines $x_4$ (just solve for it), hence the dimension of the subspace is 3. To check if what you have is a basis, (1) verify there are exactly as many vectors as dimensions (check!); (2) verify each vector is in the space (check that they all satisfy the equation), and; (3) verify that they are linearly independent (I believe you can do this by computing the Grammian determinant, IIRC...). – anon Nov 9 '11 at 23:34
The 3-dimensionality refers to the fact that once 3 vectors are known, the fourth one is determined uniquely. This object "lives" in 4-space, but it is 3-dimensional, in that it is fully describable using 3 coordinates alone. – gary Nov 9 '11 at 23:34
No. They do not form a basis because the last vector does not lie on that plane: $(-4)+2(-2)+3(2)+6(1)\not=0$. And first of all, a basis must consist of members of the space. – Bill Cook Nov 9 '11 at 23:46
@gary: Thanks I was getting tripped up about where the dimension was coming from. The way you guys explain it makes me wonder why I wasn't seeing that! – JustcallmeDrago Nov 10 '11 at 0:16

One way to form a basis for the given subspace is like this: solve the system $x_1+2x_2+3x_3+6x+4=0$. This system has 4 unknowns and one equation, therefore three of them are variables, let's say $x_2=a,x_3=b,x_4=c$. Then $x_1=-2a-3b-6c$.
Then the subspace can be written as $$S=\{ (-2a-3b-6c,a,b,c) : a,b,c \in \Bbb{R}\}$$ Now you can write it like
$$S= \{ a(-2,1,0,0)+b(-3,0,1,0)+c(-6,0,0,1) : a,b,c \in \Bbb{R}\}$$
Then $(-2,1,0,0),(-3,0,1,0),(-6,0,0,1)$ form a basis for this subspace.