Sum of subspaces Let $V_1$ be the subspace of the solutions of $x_1+x_2+x_3-x_4=0$, $V_2$ be spanned by $(1,t,1,1), (1,0,s,1)$. For what values of $s$ and $t$ do we have $V_1+V_2=\mathbb{R}^4$?
My approach is to check the condition for inclusion $V_2 \subseteq V_1$, get $t=-1, s=0$ and then notice that if $V_2 \subsetneq V_1$ there is at least one line of vectors in $V_2$ outside $V_1$. From $\text{dim}V_1=3$ and this "at least one line of vectors" I conclude $\text{dim}(V_1+V_2)=4$, so my condition is satisfied for $t \neq -1$ or $s \neq 0$. This reasoning is somewhat informal. Is there an easy way to formalize it?
Edit:
One obious way is to put all the vectors from the bases of $V_1$ and $V_2$ into a matrix and produce a row echelon form to see the dimension and check when it is $4$. I am looking for an approach quicker than this, particularly one building on the knowledge that $V_2 \subseteq V_1$ for $t=-1, s=0$.
 A: HINT:
Let $v_1,v_2,v_3$ be the vectors that span $V_1$ (they are not hard to find). Then, put these vectors along with $v_4=(1,t,1,1)$ and $v_5=(1,0,s,1)$ in the same set $A=\left\{v_1,v_2,v_3,v_4,v_5\right\}$.
Now, you have to find dimension of $span (A)$ depending on $s$ and $t$. Then, dimension is $4$ if at least one of vectors $v_4,v_5$ is linearly independent with $v_1,v_2,v_3$.
A: Yes you are right.
This intuition is correct. And you need to formalize it in Mathematical terms.
1.A vector space is completely defined by its basis.So if we study the basis of $ V_1 $ we can understand how it is related to the $ V_2 $
2.Basis of  $ V_1 $: (1,0,0,1);(0,1,0,1) ; (0,0,1 ,1)


*Now to formalize your intuition of line  we must tell that $ V_2 $ dimension must be 1 w.r.t the basis of $ V_1$


That is you have to reject those (t,s) such that the given basis of $ V_2 $ is in the span of $ V_1 $.
This is the Mathematical form of your line notion.
So, Now  as you have noticed (1,-1,1,1) and (1,0,0,1) are in   $ V_1 $. And as (0, 1,0,0) and (0,0,1,0) are not in $ V _1 $ ( easy to see) .So your 
Condition is true .
