Property of eigenvectors in linear mapping Let $V$ be a bector space over a filed $\mathbb{F}$, and let $L:V\rightarrow V$ be a linear mapping. Let $U$ be a subspace of $V$ such that $L(U)\subset U$
Suppose that $u$ and $v$ are eigenvectors of $L$ corresponding to dinstinct eigenvalues. Show that if $u+v\in U$, then $u$ and $v$ are elements of $U$.
 A: Suppose
$$\begin{cases}Lu=\lambda u\\{}\\Lv=\mu v\end{cases}\;\;\;,\;\;\;\lambda\,,\,\mu\in\Bbb F\;,\;\;\;\lambda\neq\mu$$
Then
$$U\ni L(u+v)=Lu+Lv=\lambda u+\mu v$$
So
$$\begin{cases}&u+v=u'\in U\\{}\\&\lambda u+\mu v=u''\in U\end{cases} \implies (\lambda-\mu)u=u''-\mu u'\implies u=\frac1{\lambda-\mu}(u''-\mu u')\in U$$
and likewise for $\;v\;$ .
A: Let
$Au = \lambda u, \tag{1}$
and
$Av = \mu v, \tag{2}$
with $\lambda \ne \mu$.  Since $U$ is a subspace and $L(U) \subset U$ and $u + v \in U$, we have not only
$w_1 = \lambda u + \mu v = L(u + v) \in U, \tag{3}$
but also
$w_2 = \lambda(u + v) = \lambda u + \lambda v \in U; \tag{4}$
then
$(\mu - \lambda) v = w_1 - w_2 \in U; \tag{5}$
now since $U$ is a subspace and $\mu - \lambda \ne 0$, we have $v \in U$; essentially the same argument with the roles of $u$ and $v$ reversed yields $u \in U$ as well.  QED.
Hope this helps.  Seasonal Cheers,
and as ever,
Fiat Lux!!!
A: Since $u$ and $v$ are eigenvecotrs, $L(u)=\lambda_u u$ and $L(v)=\lambda_v v$, also since $L$ is a linear mapping, $L(u+v)=L(u)+L(v)=\lambda_u u+\lambda_v v\in U$.
Since $U$ is a subspace, any linear combination of its elements is also a member of $U$. Hence, $(\lambda_u u+\lambda_v v)-\lambda_u(u+v) = (\lambda_u - \lambda_v)v\in U$. If $(\lambda_u - \lambda_v)v\in U$, then $v\in U$ since any multiple of $(\lambda_u - \lambda_v)v$ is in $U$, so $(1/(\lambda_u - \lambda_v))[(\lambda_u - \lambda_v)v]=v\in U$.
Since $v\in U$, $(u+v)-1(v)\in U$, therefore $u\in U$
