How can one complete a set to a vector basis? What are the possible ways of solving next trivial task:
$$
\mathbf{u} =
\left( \begin{array}{c}
1 \\
2 \\
0 \\
\end{array} \right)
\mathbf{v} =
\left( \begin{array}{c}
5 \\
5 \\
2 \\
\end{array} \right) \\ $$
Complete the set $\{u,v\}$ to a vector basis of the vector space $\mathbb{R^3}$
 A: You're in $\mathbb{R}^3$, so there is a way which allows to do that without thinking. If $u$ and $v$ are linearly independent, then the cross product 
$$
w:=u\times v
$$
automatically gives you a basis $(u,v,w)$.
Of course, with some little experience, you see at once that $w=(1,0,0)$ works also without any computation. For instance, the matrix with columns $(w,u,v)$ is then upper triangular. So the determinant is the product of the diagonal coefficients: $1\cdot 2\cdot 5=10\neq 0$. This proves that $(w,u,v)$ is a basis.
A: Step 1: Pick any vector for the third vector.
Congratulations; if you haven't done something silly (like pick $\vec{0}$ or $\vec{u}$), you almost certainly have a basis!
Step 2: Check that you have a basis.
If you have bad luck and this check fails, go back to step 1.
A: Hint: $\{u,v,w\}$ will be a basis of $\mathbb{R}^3$ if and only if 
$$\det\left(\begin{array}{c|c|c} & & \\ u & v & w\\ & & \end{array}\right)\ne 0$$
this gives you an inequation for $w$. 
A: Take the standard basis $e_1,e_2,e_3$. Try $\{u,v,e_k\}$. At least one of them will work.
(What I mean is pick $k$, and check if $e_k \in \operatorname{sp} \{u,v\}$. If not, you are finished, if so, pick another $k$ and repeat. Since $\dim \operatorname{sp} \{u,v\} = 2$, and $\dim \operatorname{sp} \{e_k\} = 3$, at least one of them will work.)
A: You need any three vectors in $\mathbb R^3$ that are not linearly dependent.  If you know about upper triangular matrices, I would just pick $\mathbf{w}=\begin {pmatrix} 1\\0\\0 \end {pmatrix}$.  Now you just need to show the set is linearly independent.  If we have $a\mathbf u + b\mathbf v+c\mathbf w=0$, first you need $c=0,$ then $b=0$, then $a=0.$
