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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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    @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!2011-11-10

1 Answers 1

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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.