Eigenspaces are in direct sum Let $V$ a vector space of finite dimension. Let $E_1,...,E_n$ Eigenspace associate to the eigen value $\lambda _1,...,\lambda _n$. I want to prove that $$E_1+\cdots+E_n=E_1\oplus\cdots\oplus E_n.$$
Let $v_1\in E_1,...,v_n\in E_n$ s.t. $v_1+\cdots +v_n=0$. I have to prove that $v_i=0$ for all $i$. I know that if $v_1\in E_1,...,v_n\in E_n$ are non zero vector, then they are free. So if I suppose that there is $v_i\neq 0$ (suppose WLOG $v_1=0$), then $$v_1=-v_2-...-v_n,$$
and thus, there is at least an other vector (let say $v_2$) that is non zero. Therefore $v_1=-v_2$ which is a contradiction. 
Question 1 : Is my proof working ? If not, what's wrong ?
Question 2 : I find my proof not elegant at all. Is there a more elegant proof ? 
 A: The flaw in your proof is that you assumed for the sake of contradiction that the vectors $v_1, \dots, v_n$ are linearly dependent ($v_1 + \dots + v_n = 0$), and used it to derive the consequence that the vectors are linearly dependent ($v_1 = -v_2$).  So you didn't really prove anything.
When reviewing your own proofs, you can ask yourself where you used each of the hypotheses given.  For instance, as other commenters have pointed out, you didn't use at all the fact that the vectors $v_i$ are eigenvectors for some linear transformation.  That's a red flag that you're skipping something important.
Like xarles says, the proof comes down to the essential fact that eigenvectors of a linear transformation corresponding to distinct eigenvalues are linearly independent.  Can you show that?
A: Q1: If you know that $(v_j)_1^n$ are independent, then the direct sum decomposition holds naturally, because now the expression of $0$ as a sum of vectors from $E_j$ would be unique, then by definition the sum is a direct sum. If you want to prove the decomposition from the square one, you might use my answer as a reference.  
Q2: I could give a proof.
We assume that $\{\lambda_j\}_1^n$ are distinct eigenvalues of a linear operator $\mathcal T \in \mathcal L(V)$.
Proof.$\blacktriangleleft$ Suppose $v_j \in E_j$ satisfy that 
$v_1 + v_2 + \cdots + v_n =0$. By definition, $\mathcal T - \lambda_j \mathcal I$ is zero mapping on $E_j$. Therefore apply $\mathcal T - \lambda_1 \mathcal I$ to $\sum_1^n v_j = 0$ yields
$$
(\lambda_2 - \lambda_1) v_2 + (\lambda_3 - \lambda_1) v_3 + \cdots + (\lambda_n - \lambda_1) v_n = 0.
$$
Now apply $\mathcal T - \lambda_2 \mathcal I$ to it and obtain
$$
\sum_3^n (\lambda_j - \lambda_2) (\lambda_ j - \lambda_1) v_j = 0.
$$
Repeatedly we could know that if we apply 
$$
(\mathcal T - \lambda_{n-1} \mathcal I)(\mathcal T -\lambda_{n-2} \mathcal I) \cdots (\mathcal T - \lambda_1 \mathcal I)
$$
to 
$$
v_1 + v_2 + \cdots + v_n = 0,
$$ 
then we obtain 
$$
\prod_{j=1}^{n-1} (\lambda_n - \lambda_j) v_n = 0.
$$
Since all $\lambda_j$ are distinct, $v_n = 0$.
Similarly, apply
$$
\prod_{j \neq k}(\mathcal T -\lambda_j\mathcal I) \quad [k = 1,2, \ldots, n-1]
$$
to $v_1 + \cdots + v_n =0 $ would yield similar expression
$$
\prod_{j \neq k} (\lambda_k - \lambda_j) v_k = 0,
$$
hence $v_k = 0$.
Conclusively, $v_j = 0$ for all $j$, as we desired. $\blacktriangleright$
