7.Let $V$ be a finite dimensional vector space over $K$, and let $A:V\to V$ be a linear map. Let $v\in V$ be an eigenvector of $A$, say $Av=\lambda v$. If $f$ is a polynomial in $K[t]$, show that $f(A)v=f(\lambda)v$.Linear Algebra, Serge Lang.
It is true that $A-\lambda I=0$, so that $f(A-\lambda I)=0$,since $f(A-\lambda I)=a_n(A-\lambda I)+…+a_0 (I-I)=0$. However I cannot prove that $f(A)-f(\lambda I)=0\implies f(A)=f(\lambda I)$. I tried to use the following theorem:
Theorem: Let $f,g$ be polynomials such that $f(t)=g(t)$ for all $t\in K$. Write
$f(t)=a_nt^n+…+a_0\\g(t)=b_nt^n+…+b_0$
Then $a_i=b_i$ for all $i$.
Since its proof lies on the limit:
However the theorem assumes already that $f(t)=g(t)$, and the fact t is common to both polynomials certainly does not imply its equality.
Questions:
How can I prove the statement? Which theorem do I need? Can someone provide me a proof?
Thanks in advance|
Best Answer
Sometimes things are easy: For $f(t)=\sum_{k=0}^n a_kt^k$ we have $$f(A)v=\Big(\sum_{k=0}^n a_kA^k\Big)v=\sum_{k=0}^n a_k A^k v=\sum_{k=0}^n a_k\lambda^k v=\Big(\sum_{k=0}^n a_k\lambda^k\Big)v=f(\lambda)v$$