I believe that for a linear space (assume finite first, but I'd like to hear about the case for infinite dimensional space.)
$$V_1 + V_2 = V_3$$
does not imply that
$$V_1 = V_3 – V_2.$$
Simply consider the case that $V_1$ is spanned by $\{e_1, e_2\}$, but $V_2$ is spanned by $\{e_2, e_3\}$, They sum up to $V_3 = \mathbb{R}^3$, but $V_3 – V_1 = e_3 \neq V_2$.
Therefore, operations on linear space does not obey basic arithmetic operations, right?
Why this question?
I was working on my bonus question The proof in textbook on The Transversality Theorem.
So from transversality theorem, we have
$$\operatorname{im} df_x T_x(X) + T_z(Z) = T_y(Y)$$
I was trying to make this question independent from that post, so I replaced with vector spaces $V_1, V_2, V_3$.
In Andreas Blass's great answer for this question, How to show $\dim T_x(X) = \dim df_x T_x(X).$, he proved $\dim T_x(X) = \dim df_x T_x(X)$ by transversality
$$\dim f_xT_x(X)\geq \dim Y-\dim Z$$
and rank-nullity that
$$\dim f_xT_x(X)\leq\dim X.$$
But they I start to wonder that transversality may be enough by its own, since dimensional complementarity of $X,Z$ to $Y$ is given:
$$\operatorname{im} df_x T_x(X) + T_z(Z) = T_y(Y) \Rightarrow \operatorname{im} df_x T_x(X) = T_y(Y) – T_z(Z) = T_x(X).$$
Hence $\dim T_x(X) = \dim df_x T_x(X)$.
This "shortcut" is wrong since the minus operation is not granted for vector space. So Andreas blass' original approach with inequalities is correct and no shortcut known. X-)
Best Answer
To have "arithmetical operations" you need to define the symbol "$-$" to mean the additive inverse of "$+$".
For this to be possible, you need your addition to satisfy cancellation, i.e. $x+z=y+z$ should imply $x=y$. This is not the case with you "addition": using your own example, $$ V_1+V_2=\mathbb R^3+V_2, $$ but you cannot conclude that $V_1=\mathbb R^3$.
The notation $V_1-V_2$ is used from time to time to mean the set $\{v_1-v_2:\ v_1\in V_1,\ v_2\in V_2\}$ but, as was already mentioned in the comments, for subspaces it is the same as $V_1+V_2$. The point is that this idea is meant to be used with subsets and not with subspaces.
The set difference, as you also mention, could also be denoted with the symbol "$-$", but in this case the "difference" will usually not even be a subspace.