Find a vector in the matching dimension that is not in the span I have the following vector $(1,2,-2),(2,-1,1)$. How do I find a vector that is not in the span of those two vectors. I can pick an arbitrary third vector and make the other two vectors equal to it but that will be time consuming and will most likely have to pick many other vectors. Is there a quick way to answer this?
 A: If you reduce the matrix $\begin{bmatrix}1&2&-2\\2&-1&1\end{bmatrix}$ to reduced row echelon form $\begin{bmatrix}1&0&0\\0&1&-1\end{bmatrix}$,
solving for the nullspace gives $x=0, y=t, z=t$; so $(0,1,1)$ is a basis for the nullspace.  
Since the nullspace is the orthogonal complement of the row space, $(0,1,1)$ is a vector not in the span of the given vectors.

Alternate solution:
Reduce the matrix $\begin{bmatrix}1&2&x\\2&-1&y\\-2&1&z\end{bmatrix}$ to $\begin{bmatrix}1&2&x\\0&-5&y-2x\\0&0&y+z\end{bmatrix}$; 
this shows that a vector $(x,y,z)$ is in the span of the two given vectors iff $y=-z$.
A: Pick an arbitrary vector (such as (1,0,0)), put the vectors as the rows of a matrix and take the determinant of the matrix. If the determinant is not zero then your choice is independent of the two given vectors.
A: If you know about cross products, you're done: just take $w = v_1 \times v_2$, where the $v_i$ are your two vectors. 
If not, think of the following: the span of the two vectors is a plane through the origin in 3-space. It can therefore contain at most two of $(1,0,0)$, $(0, 1, 0)$ and $(0, 0, 1)$. That means one of these three will suffice as an answer. You'll need ot show it's linearly independent from the other two, but that's not a lot of work. 
A third possibility: Choose randomly. The probability that you happen to pick a vector in the plane spanned by the other two is negligible, so your "a lot of work" amounts to "probably very little work." 
