Proof that the sum of two subspaces of $\mathbb{R}^n$ spans the entire $\mathbb{R}^n$. Is this a valid proof? I was looking at the question: Linear Algebra Subspaces Proving and I didn't really understand the proposed solutions. I have a solution of my own which I would like verified.
The question is as follows:
Let $V_1$ and $V_2$ be subspaces of $\mathbb R^n$ defined by
$$
V_1 = \{(x_1, x_2, ..., x_n) | x_1 + x_2 + \cdots + x_n = 0\}
$$
$$
V_2 = \{(x_1, x_2, ..., x_n) |  x_1 = x_2 = \cdots = x_n\}.
$$
Prove that any vector $v \in\mathbb R^n$ can be uniquely expressed as $v = v_1 + v_2$ such that $v_1 \in V_1$ and $v_2 \in V_2$.
My proof:
Since $V_1$ has $n$ elements, but only one linear equation, we can express any vector in $V_1$ with $n - 1$ parameters, i.e., $((-x_2 + -x_3 + ... + -x_n), x_2, x_3, ..., x_n)$.
This means that $\dim(V_1) = n - 1$, since there are $n-1$ parameters.
$V_2$ can also be expressed as 
$$V_2 = \{t * (1, 1, ..., 1) | v \in\mathbb R\}$$
This implies that $V_2 \not\subseteq V_1$ since $t(1) + t(1) + ... + t(1)$ $\ne$ $0$ for all $t \in \mathbb R$, which means that every vector in $V_2$ is linearly independent to every vector in $V_1$.
Since
$$\dim(V_1) = n - 1$$ and any vector in $V_2$ is linearly independent to every vector in $V_1$,
$$\dim(V_1 + V_2) \ge \dim(V_1) + 1 \ge n$$
But since $\dim(V_1 + V_2)$ cannot be $\gt n$, $\dim(V_1 + V_2)$ must be $= n$.
Hence,
$$\text{span}(V_1 + V_2) = \mathbb R ^n$$
I'm pretty new to this so this is a pretty messy proof, but is it at all valid?

To show existence, consider the decomposition
$$(x_1, \ldots, x_n) = \underbrace{\left(x_1 - \sum_{i=1}^n x_i, \ldots, x_n - \sum_{i=1}^n x_i\right)}_{\in V_1} - \underbrace{\left(\sum_{i=1}^n x_i, \ldots, \sum_{i=1}^n x_i\right)}_{\in V_2}$$
To show uniqueness, notice that $V_1 \cap V_2 = \{0\}$. Indeed, if $$(x_1, \ldots, x_n) \in V_1 \cap V_2$$
then $x_1 = \cdots = x_n$ and $0 = \sum_{i=1}^n x_i$ so $x_1 = \cdots = x_n = 0$.
Now assume there are $v_1, v_1' \in V$ and $v_2, v_2' \in V_2$ such that $x = v_1 + v_2 = v_1' + v_2'$. Rearranging gives
$$\underbrace{v_1 - v_1'}_{\in V_1} = \underbrace{v_2' - v_2}_{\in V_2}$$
so $v_1 - v_1'$ and $v_2' - v_2$ are both elements of $V_1 \cap V_2$ and hence equal to $0$. We conclude $v_1 = v_1'$ and $v_2 = v_2'$.
