Find two vectors in span=$\{u_1,u_2,u_3\}$ that are not in span=$\{v_1,v_2\}$. Let $v_1=(1,1,-1)$, $v_2=(-1,1,1)$ and $u_1=(0,1,0)$ $u_2=(1,0,-1)$, $u_3=(1,1,1)$.
Find two vectors in span=$\{u_1,u_2,u_3\}$ that are not in span=$\{v_1,v_2\}$.
I have gotten up to the point of a RREF, but I'm stuck there somewhat. 
From this
$\left({\begin{array}{cc|c|c}
1&-1&2&3\\ 
1&1&2&1\\ 
-1&1&0&-1\end{array}}\right)$, I have reduced it through Gauss-Jordon Elimination to 
$\left({\begin{array}{cc|c|c}
1&-1&2&3\\ 
1&1&2&1\\ 
0&0&2&2
\end{array}}\right)$. 
Since the system is inconsistent, we can conclude that span=$\{u_1,u_2,u_3\}$ is not in span=$\{v_1,v_2\}$. 
Where do I go from here? 
 A: As we want to avoid vectors that are linear combinations of $v_1$ and $v_2$, let us find how those vectors look like: $a(1,1,-1) +b(-1,1,1)=(a-b,a+b,b-a)$. This shows that we have to avoid vectors $(x,y,z)$ such that $x+z=0$.
We find that $u_1,u_2$ are both vectors satisfying that condition. SO we have to avoid anything which is a combination of $u_1$ and $u_2$ alone.
The vectors of the form $au_1+ bu_3=(b,a+b,b)$ with $b\ne 0$ provide infinitely many examples you desire.
A: The vectors in$\{u_1,u_2,u_3\}$ are linear independent and so any 3 dimensional vector can be made. But the set of vectors $\{v_1,v_2,\}$ only contain two vectors, so the span of vectors in $v$ form a subset of the span from $u$. Now two find vectors that are not in the span of $\{v_1,v_2,\}$, why don't we just consider its basis elements and add them up: A new vector $(0,2,0)$. Let's add $1$ to the $z$ component to arrive at $(0,2,1)$. If $a$ and $b$ are scalars, consider $a(1,1,-1)+b(-1,1,1)=(0,2,1)$, then I leave it up to you to verify that $(0,2,1)$ is NOT in span $\{v_1,v_2,\}$, but it (obviously) is in span$\{u_1,u_2,u_3\}$, so this vector answers your question. As a result, any multiple of $(0,2,1)$ will also work (why?) So now you can work it out...
