# $f(A)v=f(\lambda)v$, polynomial of eigenvalues, proof.

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?

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

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

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

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

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.

\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}