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I was studying from Technion linear algebra course when I came upon this question:

Let $V$ be a space, $0 < \dim V < \infty, T:V\to V$, and let $U_{\lambda_1}, \dots , U_{\lambda_n}$ eigenspace corresponding to (different) eigenvalues $\lambda_1, \dots \lambda_n$. And assume $W\subset V$ invariant subspace of $T$.

Assume exists basis $B$ of $V$ such that $B$ is Jordan form of $T$.

  • Prove that exists basis $B'$ such that $B'$ is Jordan form basis of $T|_W$
  • Prove that $W=(W\cap U_{\lambda_1})\oplus\dots\oplus(W\cap U_{\lambda_n})$

On the first part, we know that exists a Jordan form basis of $T$ iff the minimal polynomial of $T$ is of form $m_T(x)=\prod(x-\lambda_i)^{k_i}$ (linear product). we will show that $m_{T|_W}(x)$ is also linear, it follows that $T|_W$ has a Jordan form.

We know by definition of minimal polynomial that $m_T(T)\equiv 0,~~ m_{T|_W}(T|_W)\equiv0$.

notice that also for any $w\in W, m_{T|_W}(T)(w)=m_{T|_W}(T|_W)(w)=0$, also since $m_T(T|_W)\ne0$ we conclude that $m_T(x)~|~ m_{T|_W}(x)$, it follows immediately that $m_{T|_W}(x)$ is linear product and have a Jordan form.

The 2nd part is tricky, I want to show two things. First is that $U_{\lambda_i} \bot U_{\lambda_k}$ for $i\ne k$, then if I can show that $\bigcup U_{\lambda_i}=W$ then I get what I want.

The problem is I don't know how to show it, I am not even sure that $U_{\lambda_i} \bot U_{\lambda_k}$ for $i\ne k$ is true, how would someone do it? And is my proof for the first one correct.

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  1. The first part is easy. $V$ is a $K$-vector space (of dimension $d$) and, according to the hypothesis about the existence of $B$, the eigenvalues of $T$ are in $K$. Now, the characteristic polynomial of $T_{|W}$ divides that of $T$; consequently, the eigenvalues of $T_{|W}$ are also in $K$ and we are done.

  2. You are wrong; in fact, the $U_{\lambda_j}$ are the generalized eigenspaces of $T$, that is to say, the $\ker((T-\lambda_jI)^d)$. Then $V=U_{\lambda_1}\oplus\cdots\oplus U_{\lambda_n}$; if $x\in W$, then $x=\sum_j u_j$ where $u_j\in U_{\lambda_j}$ and it remains to prove that $u_j\in W$.

Let $f=(T-\lambda_2I)^d\cdots(T-\lambda_nI)^d$; one has $f(x)=f(u_1)\in W\cap U_{\lambda_1}$; moreover $g=f_{|U_{\lambda_1}}$ is an isomorphism and $g^{-1}$ is a polynomial in $g$, that implies that $W\cap U_{\lambda_1}$ is $g^{-1}$-stable. Finally $u_1\in W\cap U_{\lambda_1}$ and we are done.

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  • $\begingroup$ I haven't learnt what $g^{-1}$ stable means, and "if $x\in W$, then $x=\sum_j u_j$ where $u_j\in U_{\lambda_j}$", can you explain how we prove that? $\endgroup$ – Rab Sep 6 '17 at 13:51
  • $\begingroup$ 1. stable (that is invariant) for $g^{-1}$. 2. If $x\in W$, then $x\in V$. $\endgroup$ – loup blanc Sep 6 '17 at 14:05

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