4
$\begingroup$

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 $$ \operatorname{span} \{x,(A-\lambda_iI)x,...,(A-{\lambda_i}I)^{p-1}x\}\cap \operatorname{span} \{y,(A-\lambda_iI)y,...,(A-\lambda_iI)^{p-1}y \}=\{0\}?$$

$\endgroup$

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ Actually, this question is from the first tutorial on the Jordan normal form: math.tamu.edu/~dallen/m640_03c/lectures/chapter8.pdf. In the page 232, I don't understand the first paragraph. It says how to construct linearly independent invariant subspace by using the generalized eigenvectors. In the first paragraph, it says that by this method, Each of these constructed invariant subspaces is linearly independent from the others. I don't see why this is true. And this question is related to this. $\endgroup$
    – stephenkk
    Commented May 9, 2017 at 6:23
  • 1
    $\begingroup$ Yeah, you have to be more careful about the process than that exercise suggests. One way to do it is each time you pick an invariant subspace, you also pick an invariant linear complement of it, and require all the later generalized eigenvectors you choose to be contained in the complement. $\endgroup$ Commented May 9, 2017 at 14:50
  • $\begingroup$ I have trouble understanding the same point. Could you please elaborate a bit more? I mean how do we ensure that the chains in this procedure in the notes are linearly independent. $\endgroup$
    – perlman
    Commented Oct 11, 2017 at 23:22
  • $\begingroup$ @MathewJames: Writing it out in full detail would be fairly complicated. I suggest you ask that as a new question. $\endgroup$ Commented Oct 12, 2017 at 1:40

You must log in to answer this question.

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