Trivial intersection of generalised eigenspaces

Let $\lambda_1,...,\lambda_n$ be a set of distinct eigenvalues for a linear operator $T:V \to V$. Define $$\Lambda_{\lambda_i}=\{v \in V | (\lambda_i I-T)^{n_i}(v)=0 \text{ for some } n_i\in \mathbb N\}$$ I am trying to prove that the subspaces $\Lambda_{\lambda_1},\Lambda_{\lambda_2},...,\Lambda_{\lambda_n}$ are linearly independent. My approach is to use induction on $n$.

For the base case $n=1$ the result is trivial since $\forall v\in\Lambda_{\lambda_1}$ if $v=0$ then $v=0$.

Now assume that $\Lambda_{\lambda_1},...,\Lambda_{\lambda_{r}}$ are linearly independent for some $r<n$ (Induction hypothesis). Suppose that $$v_1+ \dot{} \dot{} \dot{}+v_r+v_{r+1}=0$$ where each $v_i \in \Lambda_{\lambda_i}$ for $i=1,...,r+1.$ We can apply $(\lambda_{r+1}I-T)^{n_{r+1}}$ to both sides of the equation to get that $$(\lambda_{r+1}I-T)^{n_{r+1}}(v_1+ \dot{} \dot{} \dot{}+v_r)+(\lambda_{r+1}I-T)^{n_{r+1}}(v_{r+1})=0$$ and since $v_{r+1}\in \Lambda_{\lambda_{r+1}}$ we get that $$(\lambda_{r+1}I-T)^{n_{r+1}}(v_1+ \dot{} \dot{} \dot{}+v_r)=0$$ Now, in order to use my induction hypothesis I need to show that $v_1+ \dot{} \dot{} \dot{}+v_r=0$ and this is where I get stuck. Why must it be true at this point that $v_1+ \dot{} \dot{} \dot{}+v_r=0$?

• What does it mean for a collection of subspaces to be linearly independent? This is not standard terminology. – littleO Jan 30 '17 at 5:37
• @littleO By linearly independence I mean that a subspace $A$ is linearly independent with a subspace $B$ if $\forall a \in A$ and $b\in B$, $a$ and $b$ are linearly independent in $V$ – gene Jan 30 '17 at 5:39
• Maybe that terminology is used more often than I realize. – littleO Jan 30 '17 at 5:46
• @littleO usually we say $A \cap B = \{0\}$ – Omnomnomnom Jan 30 '17 at 6:18

We can show a more general statement: If $p_1,\ldots,p_n$ are mutually coprime polynomials, and $v_1,\ldots,v_n$ are vectors $\neq 0$ satisfying $$p_i(T)\cdot v_i = 0\;\;\forall i\in\{1,\ldots,n\}$$ then the vectors $v_1,\ldots,v_n$ are linearly independent.
Let $k\in\{1,\ldots,n\}$. Let $\bar{p}_k=p_1\cdot\ldots\cdot p_{k-1}\cdot p_{k+1}\cdot\ldots\cdot p_n$. As $p_k$ and $\bar{p}_k$ are coprime, there are polynomials $q_k$ and $\bar{q}_k$ satisfying $$q_kp_k+\bar{q}_k\bar{p}_k = 1$$ Let $B_k = q_k(T)p_k(T)$ and $\bar{B}_k = \bar{q}_k(T)\bar{p}_k(T)$. As $q_kp_k+\bar{q}_k\bar{p}_k = 1$, we have $B+\bar{B} = I$.
Now we check if there is a non-trivial solution of $\alpha_1 v_1+\ldots+\alpha_n v_n=0$.
We have $$0 = \bar{B}_k v_i \;\;\forall i\in\{1,\ldots,k-1,k+1,\ldots,n\}$$ because $\bar{B}_k$ contains $p_i(T)$ as one of its factors, and all the factors commutate. Therefore, $$0 = \bar{B}_k (\alpha_1 v_1+\ldots+\alpha_{k-1}v_{k-1}+\alpha_{k+1}v_{k+1}+\ldots+\alpha_n v_n)$$ On the other hand, we have $$0 = \bar{B}_k (\alpha_1 v_1+\ldots+\alpha_n v_n)$$ because we assumed $\alpha_1 v_1+\ldots+\alpha_n v_n=0$. If we subtract the last two equations, we get $$\bar{B}_k\alpha_k v_k = 0$$ As $B_k\alpha_k v_k = 0$, we finally find $$\alpha_k v_k = (I)\cdot\alpha_k v_k = (B_k+\bar{B}_k)\cdot\alpha_k v_k = B_k\alpha_k v_k+\bar{B}_k\alpha_k v_k = 0$$ from which we conclude $\alpha_k = 0$