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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|

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    $\begingroup$ use $A^nv=\lambda ^n v$ $\endgroup$ Aug 28, 2017 at 11:21
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    $\begingroup$ It is not true that $A-\lambda I=0$. It is true that $\det(A-\lambda I)=0$. $\endgroup$
    – user228113
    Aug 28, 2017 at 11:21
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    $\begingroup$ To prove that $$f(A)-f(\lambda I)=0\implies f(A)=f(\lambda I)$$ simply take the first equation and add $f(\lambda I)$ to both sides. $\endgroup$
    – 5xum
    Aug 28, 2017 at 11:22
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    $\begingroup$ Moreover, it is by far not true that $f(A-\lambda I)=f(A)-f(\lambda I)$. Since when does $(x+y)^5=x^5+y^5$? $\endgroup$
    – user228113
    Aug 28, 2017 at 11:24
  • $\begingroup$ I must be missing something why can't you just write out everything on the LHS explicitly? $\endgroup$
    – IAmNoOne
    Aug 28, 2017 at 11:25

3 Answers 3

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Sometimes things are easy: For $f(t)=\sum_{k=0}^n a_kt^k$ we have $$f(A)v=\Big(\sum_{k=0}^n a_kA^k\Big)v=\sum_{k=0}^n a_k A^k v=\sum_{k=0}^n a_k\lambda^k v=\Big(\sum_{k=0}^n a_k\lambda^k\Big)v=f(\lambda)v$$

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Consider $f(t)=a_nt^n+\ldots+a_0$ where $a_i\in K$ for $i=1,\ldots,n$.

Do you remember how $f(A)$ is defined for $A\in K^{n\times n}$?

$$f(A)=a_nA^n+\ldots+a_0I$$

Now consider what $f(A)v$ is for an eigenvector $v$ such that $Av=\lambda v$. You get

$$f(A)v=\left(a_nA^n+\ldots+a_0I\right)v=a_nA^nv+\ldots a_0Iv.$$

Now you can use $Av=\lambda v$ to change $A$ to $\lambda$. Consider $A^nv=\lambda^n v$ since $$A^nv=A^{n-1}Av=A^{n-1}\lambda v=\lambda A^{n-1}v=\lambda A^{n-2}Av=\ldots=\lambda^2A^{n-2}v=\ldots=\lambda^nv$$and you get

$$f(A)v=\ldots=a_n\lambda^nv+\ldots+a_0v=\left(a_n\lambda^n+\ldots+a_0\right)v=f(\lambda)v.$$

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Define $f(A) = c_0 + c_1A + c_2A^2 + \dots c_nA^n$, then $f(A)$ acting on $v$ yields $V$ as a $K[x]$ module.

For your question,

\begin{align} f(A)v &= (c_0 + c_1A + c_2A^2 + \dots c_nA^n)(v)\\ &= c_0 + c_1Av + c_2A^2v + \dots c_nA^nv\\ &=c_0 + c_1\lambda v + c_2A(\lambda v) + \dots c_nA^{n-1}(\lambda v) \quad \text{recall $A\lambda = \lambda A$}\\ &=c_0 + c_1\lambda v + c_2(\lambda^2 v) + \dots c_n(\lambda^n v) \\ &=(c_0 + c_1\lambda + c_2\lambda^2 + \dots c_n\lambda^n)(v)\\ &=f(\lambda)v \end{align}

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