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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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.

  • $\begingroup$ 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

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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