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Let $$T:V\to V$$ be a linear transformation. Suppose that $v_1, v_2, \cdots, v_k \in V$ are eigenvectors of $T$ that correspond to distinct eigenvalues. Assume that $W$ is a $T$-invariant subspace of $V$ that contains the vector $v_1 + v_2 + \cdots + v_k$. Show that $W$ contains each of $v_1, v_2, \cdots, v_k$.

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    $\begingroup$ Welcome to math.SE. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. $\endgroup$ – Michael Albanese May 9 '15 at 22:00
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Let $\lambda_i$ be the eigenvalue associated to $v_i$.

Proof by induction. For $k=1$ this is trivial. Now assume that $v_1+...+v_k\in W$. But then also $\lambda_1v_1+...+\lambda_kv_k\in W$ by $T$-invariance and $\lambda_1v_1+...+\lambda_1v_k\in W$. Hence $(\lambda_2-\lambda_1)v_2+...+(\lambda_k-\lambda_1)v_k\in W$. Now apply the induction hypothesis on $(\lambda_2-\lambda_1)v_2,...,(\lambda_k-\lambda_1)v_k$ and receive $v_2,...,v_k\in W$. But then of course $v_1\in W$ aswell.

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  • $\begingroup$ I am confused on how we know that λ1v1+...+λ1vk∈W ... How can we deduce that λ1 acts as an eigenvalue for each of the eigenvectors? I thought that λ1 was the eigenvalue corresponding to precisely v1. I am also confused on how applying the induction hypothesis allows us to arrive at our conclusion. @LeBtz $\endgroup$ – quantumlogic May 9 '15 at 22:49
  • $\begingroup$ $W$ is a subspace, so for each scalar $\mu$ and each $x\in W$ we have $\mu x\in W$. Apply this on $x = v_1+...+v_k$ and $\mu = \lambda_1$, it's really basic. What exactly do you not understand concerning the use of induction hypothesis? $\endgroup$ – Tim B. May 9 '15 at 22:53
  • $\begingroup$ That makes sense. How does assuming that v1+...+vk∈W allows us to conclude that v2,...,vk∈W ? @LeBtz $\endgroup$ – quantumlogic May 9 '15 at 23:01
  • $\begingroup$ This is not the induction hypothesis. The induction hypothesis states: For each $l<k$ and for each set of eigenvectors $w_1,...,w_l$ with distinct eigenvalues, if $w_1+...+w_l\in W$ then $w_1,...,w_l\in W$. In this case $l = k-1, w_1 = (\lambda_2-\lambda_1)v_2$ and so on. $\endgroup$ – Tim B. May 9 '15 at 23:03
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Let $v=v_1+\ldots+v_k$, $T(v_i)=\lambda_iv_i$, $f_i(x)=x-\lambda_i$ for all possible $i$. Since $\lambda_i$ are all distinct, then $\gcd(f_i(x),f_j(x))=1$ for $i\neq j$. By the chinese reminder theorem, there exists polinomials $p_i(x)$ such, that $p_i\equiv \delta_{i,j}\pmod{f_j}$, where $\delta_{i,j}$ - Kronecker symbol. Then $p_i(T)(v_j)=\delta_{i,j}v_i$, so $p_i(T)(v)=v_i$. Since $W$ is a $T$-invariant and $v\in W$, then $v_i\in W$ for all $i$.

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