Extending basis of a vector subspace I was given the equation of the plane $x-y+z =0$. I have already shown that this is a vector subspace. From this I have $z = y-x$. From this I found a basis. Namely $<(1,1,0), (1,0,-1)>$. My book is asking me to extend this to $\Bbb R^3$. I don't really know how to extend the basis as it is asking.
 A: Take the cross product of the two vectors you found, and add it to the bunch.
A: A linear combination  of your choice of basis for the plane is $a(1,1,0)+b(1,0,-1)=(a+b,a,-b)$. Now choose a vector not in the form. Clearly $(1,0,0)$ is not in this form and so is not a linear combination. So the set
$\{(1,1,0),(1,0,-1), (1,0,0)\}$ is an extended basis.
A: Finding an orthogonal vector via cross product works great in $\mathbb R^3$. If you’re working in a space that’s not of dimension 3, the same basic idea works, but you’ll have to approach it a little differently. You still look for the orthogonal complement of the subspace, but since you can’t use a cross product, instead form the matrix $\begin{bmatrix}v_1&\cdots&v_m\end{bmatrix}^T$, where the $v_k$ are the basis for the subspace (i.e., form the matrix with these vectors as its rows) and find its nullspace (kernel).
In this case, you’d look for the kernel of $$\begin{bmatrix}1&1&0\\1&0&-1\end{bmatrix}$$ which you can find to be the span of $(1,-1,1)^T$ via row-reduction. For this simple case, you can probably find this vector via inspection—you’re looking for a vector whose dot products with both of the plane’s basis vectors is zero.
