0

I am trying to understand how do we form a Jordan Normal Basis and more specifically why the chains are linearly independent. I saw some relevant question, but still, I can't figure that out.

We start with a highest order eigenvector and form its Jordan chain. Then, if we haven't had yet a basis, we take another eigenvector of highest possible order and form its Jordan chain and we continue like this. Why will two jordan chains we obtain be linearly independent?

Thanks in advance

edit:In this post, he says:

The other method is to first determine the lengths of the chains (there are formulas you can use to find them). If the longest chains have length kk, then you find linearly independent vectors in $(T−\lambda I)^{k−1}$ that are not in $(T−\lambda I)^{k−2}$. Then you apply $(T−\lambda I)$ to them, and if necessary find linearly independent vectors from the ones you have so far that are in $(T−\lambda I)^{k−2}$ but not in $(T−\lambda I)^{k−3}$. You continue this way, always selecting linearly independent vectors, until you are down to finding enough linearly independent vectors in $(T−\lambda I)$ besides the ones you already have; you will be guaranteed they are linearly independent because at each stage we are always selecting vectors that are linearly independent.

My question is actually, how do we know that after applying $(T−\lambda I)$ to the vectors we already have selected, we get a linearly independent set of vectors?

perlman
  • 241
  • 2
  • 11
  • For simplicity, let $x$ be a top level vector such that $A^2x = 0$. Now choose another top level vector $y$ and assume $Ay = 0$. You ask why $x,Ax$ and $y$ are linearly independent. So, let $ax+bAx+cy = 0$. Applying $A$ yields $aAx = 0$ and hence $a=0$. Now, $bAx = -cy$. If $c\neq 0$, then also $b\neq 0$ because $y\neq 0$. But then $y = -(b/c)Ax\in im,A$ and thus $y$ is not a top level vector. Hence, $b=c=0$. – amsmath Oct 11 '17 at 23:56
  • @amsmath If $y$ was such that $A^2y=0$ as well, then we would end up with $aAx+bAy=0$. How do we get a contradiction in this case? – perlman Oct 12 '17 at 00:15

1 Answers1

3

The linked answer is slightly imprecise; this part

are not in $\ker(T-\lambda I)^{k-2}$

should read

span a subspace of $\ker(T-\lambda I)^{k-1}$ complementary to $\ker(T-\lambda I)^{k-2}$

Here is the full procedure in more detail. For convenience let $V_l=\ker(T-\lambda I)^l$. Note that $V_0=0$. Let $k$ be minimal such that $V_k=V_{k+1}$. For $l=k+1,k,\ldots,1$ we will inductively construct a linearly independent set $B_l$ spanning a subspace of $V_l$ complementary to $V_{l-1}$ (equivalently $B_l$ maps to a basis of $V_l/V_{l-1}$).

Since $V_k=V_{k+1}$, we can take $B_{k+1}=\emptyset$. Suppose $1\leq l\leq k$ and we have already constructed $$ B_{l+1}=\{v_1,\ldots,v_m\}. $$ I claim that the vectors $(T-\lambda I)v_i$ are not only linearly independent, their span has trivial intersection with $V_{l-1}$. Indeed suppose $$ \sum_i a_i(T-\lambda I)v_i\in V_{l-1}. $$ Then $$ 0=(T-\lambda I)^{l-1}\sum_i a_i(T-\lambda I)v_i =(T-\lambda I)^l\sum_i a_iv_i. $$ Thus $\sum_i a_iv_i\in V_l$. By the inductive assumption, $a_i=0$ for each $i$ as required. Thus we can obtain $B_l$ by extending the set $\{(T-\lambda I)v_i\}$ to span a subspace of $V_l$ complementary to $V_{l-1}$.

stewbasic
  • 6,131