# Are two invariant subspaces generated by two linearly independent generalized eigenvectors linearly independent?

I am learning now the Jordan normal form in linear algebra or abstract algebra. $C$ is the complex number set. Let $A\in M_n(C)$ with spectrum $\sigma(A)=\{\lambda_1,\lambda_2,...,\lambda_k\}$,and with the invariant subspaces $V_{\lambda_{i}}$. The definition of $V_{\lambda_i}$ is: $$V_{\lambda_{i}}:=\{x\in C^n|(A-\lambda_iI)^n x=0\}$$. Then it is a theorem that for any $x\in V_{\lambda_{i}}$ be a generalized eigen vector of order $p$($p$ is a positive integer indeed), i.e., a $x$ such that $(A-\lambda_iI)^px=0$ and $(A-\lambda_iI)^{p-1}x\neq0$ then $$x, (A-\lambda_iI)x,...,(A-\lambda_iI)^{p-1}x$$ spans a invariant subspace and all above vectors are linearly independent. My question is that: if there are two generalized eigen vectors $x$ and $y$. $y$ is not in the subspace generated by $x$ and with the same order $p$, then are the subspace generated by $x$ and subspace generated by $y$ linearly independent? To be more precise. is $$\operatorname{span} \{x,(A-\lambda_iI)x,...,(A-{\lambda_i}I)^{p-1}x\}\cap \operatorname{span} \{y,(A-\lambda_iI)y,...,(A-\lambda_iI)^{p-1}y \}=\{0\}?$$

No, because (for instance) $y$ might be $x$ plus an element of lower order. For an explicit example, consider the matrix $$A=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}$$ whose only eigenvalue is $\lambda=0$. The vectors $x=(0,0,1)$ and $y=(1,0,1)$ both have order $2$ and neither is in the invariant subspace generated by the other. However, $Ax=Ay=(0,1,0)$ is in the invariant subspace generated by both of them.