Only 'atomic' vectors as part of the base of a vector space? 
Given a vector subspace    $U_1=${$\left(\begin{array}{c}
        \lambda+µ \\
        \lambda \\ µ
      \end{array}\right)\in R^3$: $\lambda,µ \in R$ } 
Determine a possible base of this vector subspace.

As far as I know, the base of a vector space is a number of vectors from which every vector in the given vector (sub-)space can be constructed. 
My suggestion for a base of the given vectorspace would be:
$$\left(\begin{array}{c}
       \lambda \\
       0 \\ 0
     \end{array}\right)+
\left(\begin{array}{c}
       µ \\
       0 \\ 0
     \end{array}\right)+
\left(\begin{array}{c}
       0 \\
       \lambda \\ 0
     \end{array}\right)+
\left(\begin{array}{c}
       0 \\
       0 \\ µ
     \end{array}\right)
$$
None of these vectors can be written as a linear combination of  the other vectors. 
What also came to my mind as a possible solution was:
$$\left(\begin{array}{c}
       \lambda+µ \\
       0 \\ 0
     \end{array}\right)+
\left(\begin{array}{c}
       0 \\
       \lambda \\ 0
     \end{array}\right)+
\left(\begin{array}{c}
       0 \\
       0 \\ µ
     \end{array}\right)
$$
Which - if any - of these two is a valid base for the given vectorspace? Is the second one invalid, because it can be written like the first one?

The first thing you need to know is that a subspace's dimension cannot exceed the containing space's dimension. Since the number of vectors constituting a base is equal to the dimension, your first suggestion is wrong, as it suggests that the subspace is of dimension 4 in $\mathbb{R}^3$, which is only of dimension 3.
Then, if a subspace's dimension is equal to the dimension of the containing space, they are equal. This means that if your second suggestion is correct, the subspace $U_1$ is equal to $\mathbb{R}^3$, which is also false (take for instance the vector $(1,0,0)$)
As a general rule, you need to factor the scalars appearing in the definition (here $\lambda$ and $\mu$), like Listing suggested, and the basis will naturally appear.
