$W$ is a $T$ invariant subspace of $V$ , and we know that $\sum^{k}_{i=1}a_i \in W$. show that $\forall i, a_i \in W ( 1\leq i\leq k)$

Question

I tried to prove this with no success:

$T$ is is linear transformation from $V$ to $V$ above $\mathbb F$. $a_1,a_2....a_k$ are eigenvectors for different eigenvalues $\lambda_1,....\lambda_k$. $W$ is a $T$ invariant subspace of $V$ , and we know that $\sum^{k}_{i=1}a_i \in W$. show that $\forall i,\, a_i \in W ( 1\leq i\leq k)$

I tried to show by induction in the following way:

induction assumption: $\sum^{k-1}_{i=1}a_i \in W$ , so that means $\forall i,\, a_i \in W ( 1\leq i\leq k-1)$.

now we will show that if $\sum^{k}_{i=1}a_i \in W$ so $\forall i, a_i \in W ( 1\leq i\leq k)$ I got to the point where I show that $(\lambda_1-\lambda_k)a_1+....(\lambda_{k-1}-\lambda_k)a_{k-1}\in W$ but I cant use my induction assumption.

what is the right way?

Answer 1

Your approach is fine. The statement is obvious if $k=1$. Assume now that it holds for a certain $k$. You want to prove that if $\sum_{i=1}^{k+1}a_i\in W$, then each $a_i$ belong to $W$. Consider$$\lambda_{k+1}\left(\sum_{i=1}^{k+1}a_i\right)-T\left(\sum_{i=1}^{k+1}a_i\right).\tag1$$It belongs to $W$, of course. But $(1)$ is equal to$$\sum_{i=1}^{k+1}(\lambda_{k+1}-\lambda_i)a_i=\sum_{i=1}^k(\lambda_{k+1}-\lambda_i)a_i.$$And now you can apply the induction hypothesis.