Any operator on a nonzero finite-dimensional vector space over C has at least one
eigenvalue. Show that the finite-dimensionality is a necessary hypothesis by verifying that the operator T <- L(P (C)) given by T p(x) = xp(x) has no eigenvalues, where P means polynomial and C is the complex number.
vector space and eigenvalue
2013-07-27 2:36 pm
回答 (1)
2013-08-01 4:24 am
✔ 最佳答案
I think the answer to this question is trivial
Suppose p(x) is a polynomial of degree n
T p(x)=λp(x)
=>xp(x)=λp(x)
LHS is of degree n+1 and RHS is of degree n
The equality will never hold.
收錄日期: 2021-04-23 23:25:23
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