Question

Let
TEL(V)
, and let be a polynomial. Show that if $\lambda$ is an eigenvalue of , then $p(\lambda)$ is an eigenvalue of .

Hint: this follows from the more precise statement that if $v \in V$ is a non-zero eigenvector for for the eigenvalue $\lambda$, then is also an eigenvector for for the eigenvalue $p(\lambda)$. Prove this.

Answer 1

and PEP (f) be a cost o out a let TELCV) polynomial. To share that ist is eigenvalue of T, then P (d) is eigenvalue of PCD). proof consider apolynomial prx) E P CH) consider P (d) = a b consider P(x) = a also a polynomial over F by using fundamental theorem of Algebra We can factor prxd-a into limear factors over . le. pcad a = an (x-2) 6x-22) (x-3) -..(2-8h) and by definition of eigenvalue, there is some reto sit. 1000 Toproove (PCT)-a I) re = an CT-2) (T-82) ...(T25) (4) let i be the first facter that image of l. then (T-ri) (W) =0 for wto so ri is an eigenvalue of I Also rais a root of prach a so peri) a=0

le proi)=a Color Thus dari is the desired eigenvalue =) p (2, ') is eigenvalue of P (T) P() is eigenvalue of PGT) Hence proof!