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?
 A: There are only small errors, but all else is fine.


*

*Rather than show that $V_2 \nsubseteq V_1$, show that $V_2 \cap V_1 = \{0\}$, which is easier : If $(t , t , ... , t) \in V_1 \cap V_2$ then $nt = 0$ so $t = 0$. The error you've made there is $t(1) + t(1) + ... + t(1) = 0$ can never happen , which is not true because $t = 0$ is possible.

*That $\dim V_1 = n-1$ needs to be proven more rigorously. Fair enough, it seems that we can express any vector in $V_1$ with $n-1$ vectors, but why not lesser? The reason is that you can find $n-1$ linearly independent vectors in $V_1$, for example $(-1,1,0,...,0), (-1,0,1,0,...,0), (-1,0,0,1,0,...,0),... (-1,0,0,...,0,1)$ are linearly independent and belong in $V_1$, so that shows that the dimension is at least $n-1$. What you have shown is that these vectors span $V_1$ in your expression so you get the dimension(Also, I am sure $(-x_2,...,-x_n)$ is a typo for $-x_2-...-x_n$).
I think the rest is fine : note that $V_1 \cap V_2 = \{0\}$ and $\dim V_1 + \dim V_2 = n$ implies the statement "every $v \in V$ can be written uniquely as $v = v_1 + v_2$ with $v_i \in V_i$, $i=1,2$" but you have not shown this in the explanation. I suspect you must already know the proof of that, but if you do not then attempt it.
A: We know that $\mathbb{R}^n$ is a Hilbert space $\mathfrak H$, and we have that if $F$ is a closed linear subspace of $\mathfrak H$ we can write
$$\mathfrak H=F\oplus F^{\perp}$$
where $F^{\perp}=\{h\in\mathfrak H:\langle h,f\rangle=0,\forall f\in F\}$.
If we take $F=V_2$, for any $v_2=t(1,\cdots,1)\in V_2$ and $v_1\in V_1$
$$\langle v_2,v_1\rangle=t\sum_{j=1}^{n}v_{1,j}=0$$ 
where $v_{1,j}$ denotes the $j$-th component of the vector.
This shows that $V_1$ and $V_2$ and orthogonal, this is $V_2=F^{\perp}$.
(Notice furthermore that $V_1\cap V_2=\{0\}$)
Hence we can write that
$$\mathbb{R}^n=V_1\oplus V_2$$
A: 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 $v_2 = v_2'$.
A: The intersection $V_1\cap V_2=\{0\}$, because a vector in it has to satisfy
$$
x_1+x_2+\dots+x_n=0,\qquad x_1=x_2=\dots=x_n
$$
The subspace $V_1$ is the kernel of the nonzero map $\varphi\colon\mathbb{R}^n\to\mathbb{R}$, $\varphi(x_1,\dots,x_n)=x_1+\dots+x_n$, so it satisfies $\dim V_1=n-1$.
Since $V_2\ne\{0\}$, we have $\dim V_2\ge1$. By Grassmann's formula
$$
\dim(V_1+V_2)=\dim V_1+\dim V_2+\dim(V_1\cap V_2)\ge n+1-0=n
$$
Therefore $V_1+V_2=\mathbb{R}^n$ (which also proves that $\dim V_2=1$).
