invariant subspace and quotient space

Question

let $T \in\mathcal{L}(V)$ be a linear operator on $V$ and let $U$ be $T$-invariant subspace of $V$.
Suppose that $v_{1},...,v_{k}$ are elements of $V$ such that $$Tv_{j}=\lambda_{j}v_{j},\space \space j=1,...,k,$$ where the $\lambda_{j}$ are distinct elements of $F$. If $$(v_{1}+U)+(v_{2}+U)+...+(v_{k}+U)=0+U\in V/U,$$ show that $v_{j}+U=U$ for all $j=1,...,k$.

Answer 1

Note that $v_{j} + U = U$ for all $j =1, \cdots, k$ if and only if $v_{j} \in U$. We induct on $k$.

Base case: $k = 1$. Then by the hypothesis, $(v_{1} + U) = 0 + U$ and the result follows.

Suppose the result holds for $k-1$. Consider $k$ eigenvectors:The hypothesis implies that $v= v_{1} + \cdots + v_{k} \in U$. Since $U$ is $T$-invariant we know that $T(v) = \lambda_{1}v_{1} + \cdots + \lambda_{k}v_{k} \in U$. Note that $\lambda_{1}v - Tv \in U$. Hence, $(\lambda_{1} - \lambda_{2})v_{2} + \cdots + (\lambda_{1} - \lambda_{k})v_{k} + U = U$. By the inductive hypothesis, $v_{k} \in U$ and this completes the proof.

Answer 2

$(v_1+U)+\dots+(v_k+U)=0+U$ is equivalent to $v_1+\dots+v_k\in U$. Since $U$ is $T$-invariant then $T(v_1+\dots+v_k)\in T(U)\subset U$ i.e. $$\lambda_1v_1+\dots+\lambda_kv_k\in U$$ If we apply $T$ $i$ times, then we will get $$\lambda_1^iv_1+\dots+\lambda_k^iv_k\in U$$ Then we can see that, $v_1+U=U$ is equivalent to show that $v_1\in U$. But, we can show that there exists a non-zero linear combination $$v_1=c_1(v_1+\dots+v_k)+c_2(\lambda_1v_1+\dots+\lambda_{k-1}v_k)+\dots+c_{k-1}(\lambda_1^{k-1}v_1+\dots+\lambda_k^{k-1}v_k)\in U (*)$$ Indeed, it follows from the fact that all $\lambda_i$ are distinct and the determinant of Vandermonde matrix of distinct elements $\lambda_i$ is non-zero. The Vandermonde matrix is going to show up since $(*)$ is equivalent to $$c_1+c_2\lambda_1+\dots+c_k\lambda_1^{k-1}=1$$ $$c_1+c_2\lambda_2+\dots+c_k\lambda_2^{k-1}=0$$ $$\dots$$ $$c_1+c_2\lambda_{k-1}+\dots+c_k\lambda_{k-1}^{k-1}=0$$ So, we have a matrix equation $AC=e_1$ where $C$ is the column vector of $c_i$, $e_1$ is a vector $(1,0,\dots,0)^t$, and $A$ is a Vandermonde matrix. Since $A$ is invertible, then $$C=A^{-1}e_1$$ Therefore, $v_1\in U$. The proof for $v_j$ is similar.