I was looking at this problem set by Ben Elias on homological algebra, where he states the following problem :
Let $x$ be an operator acting on a finite dimensional vector space $V$ by a single Jordan block, with eigenvalue $\lambda$. Show that $\ker(x − \lambda I)$ is contained in any nonzero subspace.
Am I missing something or this is wrong? If I take a Jordan basis of $V$, $\{v, (x-\lambda I)v, \dots, (x-\lambda I)^{n-1}v\}$ then clearly the kernel is the one dimensional subspace generated by $v_n=(x-\lambda I)^{n-1}v$. But the nonzero subspace generated by $v$ doesn't contain the kernel.
From Mariano Suárez-Álvarez's comment :
I think one proof may be like this. Let $W$ be an invariant subspace. If we choose a Jordan basis $\{v_1, v_2, \dots, v_n\}$ where $v_i=(x-\lambda I)^{i-1}v_1$ and $(x-\lambda I)v_n=0$. Then it suffices to show $v_n\in W$. Let $w\in W$ be a nonzero element. Let $w=\sum_{i=1}^{n}a_iv_i$. Let $j$ be the smallest index for which $a_i\ne 0$. Now notice if $W$ is invariant under $x$ then it is also invariant under $x-\lambda I$. Then consider $(x-\lambda I)^{n-j}(w)=a_jv_n\in W$. So the proof is complete.