Condition of containing the eigenvectors of a Linear Operator over a finite dimensional vector space in a T invariant Subspace Let $T$ be a linear operator on a finite dimensional vector space $V$ and $W$ be a $T$-invariant subspace of  $V$.Suppose $v_1,v_2,...,v_n$ are the eigenvectors corresponding to distinct eigenvalues $\lambda_1,...,\lambda_n$. Suppose $v_1+v_2+...+v_n\in W$ and prove that  $v_1,v_2,...,v_n\in W$.
SOLUTION APPROACH:
I have taken an approach as $W$ is $T$ invariant and $v_1+v_2+...+v_n\in W$. This clearly implies $$T(v_1+v_2+...+v_n)=\lambda_1v_1+...+\lambda_n v_n\in W.$$
I am stuck here.I can't conclude anything else from here.
Thanks in advance.If I've done some mistake please do forgive me.
 A: $\mathit{Proof}. \blacktriangleleft$ We prove that $v_n \in W$, and the method is similar for other vectors.
Apply $\mathcal T$ to the vector repeatedly: 
$$
\mathcal T (\sum_1^n v_j) = \sum_1^n \lambda_j v_j
$$
and 
$$
(\mathcal T - \lambda_1\mathcal I)(\sum_1^n v_j) = \sum_2^n (\lambda_j - \lambda_1) v_j\in W, 
$$
since $W$ is invariant under $\mathcal T$ and $\lambda_1 \sum_1^n v_j \in W$. 
Next, 
$$
(\mathcal T - \lambda_2\mathcal I)(\mathcal T - \lambda_1\mathcal I)(\sum_1^n v_j) = \sum_3^n (\lambda_j - \lambda_1)(\lambda_j - \lambda_2) v_j \in W
$$
by a similar reason. It is not hard to see that 
$$
\prod_{n-1}^1(\mathcal T - \lambda_j \mathcal I)(\sum_1^n v_j) = \prod_1^{n-1}(\lambda_n - \lambda_j) v_n \in W.
$$
Since all eigenvalues are distinct, the coefficient in the above equation is nonzero, hence $v_n \in W$. 
Similarly we have 
$$
(\mathcal T -\lambda_{n-1}\mathcal I) \cdots \widehat{(\mathcal T - \lambda_k \mathcal I)} \cdots (\mathcal T - \lambda_1 \mathcal I) (\sum_1^n v_j) = \prod_{j\neq k} (\lambda_k - \lambda_j) v_k \in W
$$
[where the hat means that we omit this factor in the product]. Thus the conclusion follows. $\blacktriangleright$
