# Find a basis for the set of vectors in $\mathbb{R}^4$ in the subspace (hyperplane) $x_1 +x_2 + 2x_3 + x_4 = 0, x_1 + 2x_2-x_3=0$

I am studying for a test, and this is one of the practice problems.

Find a basis for the set of vectors in $\mathbb{R}^4$ in the subspace (hyperplane) $x_1 +x_2 + 2x_3 + x_4 = 0, x_1 + 2x_2-x_3=0$

Can I say that the second plane is a linear combination of the first plane, and a basis for the first plane is $\{\begin{bmatrix} 1 & 0 & 0 & 1 \end{bmatrix}, \begin{bmatrix} 0 & 1 & 0 & -1 \end{bmatrix}, \begin{bmatrix} 0 & 0 & 2 & -2 \end{bmatrix}\}$, thus it is the basis for the hyperplane (both planes) in the subspace? If not, how do I find the basis?

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Why is $(1,0,0,1)$ on the first plane? – wj32 Nov 7 '12 at 7:01
It might not be, I could be wrong. I just followed a heuristic I found online, since it doesn't seem to be in my textbook. – Grace C Nov 7 '12 at 7:08
Also, why do you say that the subspace is a hyperplane? It looks 2 dimensional to me. – wj32 Nov 7 '12 at 7:09
The actual problem itself says that in the question. – Grace C Nov 7 '12 at 7:11
Where is this problem from? – wj32 Nov 7 '12 at 7:18

I'm assuming you want to find a basis for the subspace $$S=\{(x_1,x_2,x_3,x_4) \in \mathbb{R}^4 \mid x_1+x_2+2x_3+x_4=0\;\mbox{and}\;x_1+2x_2-x_3=0\}.$$ The standard way to do this is to notice that $S$ is the kernel of the matrix $$\begin{bmatrix}1 & 1 & 2 & 1 \\ 1 & 2 & -1 & 0\end{bmatrix}.$$ Row reduce to get $$\begin{bmatrix}1 & 0 & 5 & 2 \\ 0 & 1 & -3 & -1\end{bmatrix}.$$ This tells you that a basis for $S$ is $\{(-5,3,1,0),(-2,1,0,1)\}$.
The last step comes from $I (x_1,x_2)^T + \pmatrix{5 & -2 \\ -3 & -1} (x_3,x_4)^T = 0$, which shows that when you select $x_3,x_4$, $x_1,x_2$ are completely defined. – copper.hat Nov 7 '12 at 7:31