5
$\begingroup$

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 $v_1, v_2, \cdots, v_k$.

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$ May 9, 2015 at 22:00

2 Answers 2

4
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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 $\endgroup$ May 9, 2015 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? $\endgroup$
    – Lukas Betz
    May 9, 2015 at 22:53
  • $\begingroup$ That makes sense. How does assuming that v1+...+vk∈W allows us to conclude that v2,...,vk∈W ? @LeBtz $\endgroup$ May 9, 2015 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. $\endgroup$
    – Lukas Betz
    May 9, 2015 at 23:03
1
$\begingroup$

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 $i$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .