Timeline for Bound the eigenvalue of product of matrices?

11 events

when toggle format	what		by	license	comment
Jun 12, 2017 at 5:41	vote	accept	CommunityBot
Jun 10, 2017 at 17:45	history	edited	user31317	CC BY-SA 3.0	edited body
Jun 10, 2017 at 3:29	answer	added	fedja		timeline score: 7
Jun 10, 2017 at 1:46	comment	added	fedja		It is funny: I can almost show it. Namely, I can show that there is a vector $x\in \mathbb R^n$ such that $\|M^nx\|\to\infty$ (that is easy: just take any $x$ with $\langle Hx,x\rangle<0$ and notice that if $y=x-\langle Hx,e\rangle e$ for a unit vector $e$, then $\langle Hy,y\rangle\le \langle Hx,x\rangle-\langle Hx,e\rangle^2$ and we cannot have the scalar products very small throughout the entire cycle). However, I cannot yet exclude the case of a non-trivial Jordan block with an eigenvalue of absolute value $1$ if $H$ is degenerate. Any ideas?
S Jun 9, 2017 at 17:22	history	suggested	Pietro Paparella	CC BY-SA 3.0	Improved exposition
Jun 9, 2017 at 17:03	review	Suggested edits
S Jun 9, 2017 at 17:22
Jun 9, 2017 at 11:15	history	edited	user31317	CC BY-SA 3.0	added 11 characters in body
Jun 9, 2017 at 8:33	history	edited	user31317	CC BY-SA 3.0	edited title
Jun 9, 2017 at 4:45	history	edited	user31317	CC BY-SA 3.0	edited body
Jun 9, 2017 at 4:19	comment	added	Anthony Quas		By the product, I guess you mean (1) $i=j$; and (2) the product should be taken in increasing order (as these matrices don't commute).
Jun 9, 2017 at 3:33	history	asked	user31317	CC BY-SA 3.0