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.
$\mathit{Proof}. \blacktriangleleft$ 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. $\blacktriangleright$