Show that $p(A)\vec{v}=p(\lambda)\vec{v}$ when $A\vec{v}=\lambda\vec{v}$ Suppose $p(x)=c_0+c_1x+...+c_kx^k$. Let $\lambda \in \mathbb{R}$, $\vec{v}\in \mathbb{R}^n$, $A_{n\times n}$, such that $A\vec{v}=\lambda\vec{v}$. Show that for any $p(x)$, $p(A)\vec{v}=p(\lambda)\vec{v}$.
It seems to me that $A\vec{v}=\lambda\vec{v}\Rightarrow A=\lambda$, so that the problem is trivially proven. Is this correct reasoning?
 A: It is sufficient to prove that for all positive integers $p$,
$$A^pv=\lambda^pv$$
This holds by choice of $A$ and $\lambda$ if $p=1$. Otherwise, assume it holds for $p-1$. Then
$$A^pv=A(A^{p-1}v)=A\lambda^{p-1}v=\lambda^{p-1}(Av)=\lambda^pv$$
and the result follows by induction.
A: It is not correct to say that
$A \vec v = \lambda \vec v \tag{1}$
for some $\vec v \ne 0$ implies
$A = \lambda I \tag{2}$
unless (1) holds for every $\vec v \in \Bbb R^n$.  A simple counterexample may be had by considering the $2 \times 2$ matrix $C$ given by
$C = \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end {bmatrix}, \tag{3}$
where
$\lambda_1 \ne \lambda_2; \tag{4}$
set
$\vec v_1 = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \; \;  \vec v_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix}; \tag{5}$
then
$C \vec v_1 = \lambda_1 \vec v_1, \tag{6}$
$C \vec v_2 = \lambda_2 \vec v_2; \tag{7}$
but this contradicts (2), since (2) implies (1) holds for all vectors $\vec v$, including $\vec v_1$ and $\vec v_2$.  In more detail:  if (1) implied (2), then by (6) we would have
$C \vec v_2 = \lambda_1 \vec v_2; \tag{8}$
but then by (7),
$\lambda_1 \vec v_2 = \lambda_2 \vec v_2, \tag{9}$
whence
$(\lambda_1 - \lambda_2) \vec v_2 = 0, \tag{10}$
forcing
$\vec v_2 = 0, \tag{11}$
contradicting the definition of $\vec v_2$ in (5).  (1) may hold for some vectors without holding for all.
As for showing that (1) implies
$p(A) \vec v = p(\lambda) \vec v, \tag{12}$
for $p(x) = \sum_0^k c_i x^i$,
we observe that
$A^2 \vec v = A(A \vec v) = A(\lambda \vec v) = \lambda A \vec v = \lambda^2 \vec v; \tag{13}$
furthermore if 
$A^m \vec v = \lambda^m \vec v, \tag{14}$
then
$A^{m + 1} \vec v = A^m A \vec v = A^m (\lambda \vec v) = \lambda A^m \vec v = \lambda^{m + 1} \vec v; \tag{15}$
with (1) and (2) forming the base hypotheses, (14)-(15) completes a simple inductive proof that 
$A^i \vec v = \lambda^i \vec v \tag{16}$
holds for all integers $i \ge 0$ (the $i = 0$ case being the identity $\vec v = \vec v$).  Thus
$c_iA^i \vec v = c_i \lambda^i \vec v, \tag{17}$
whence
$p(A) \vec v = \sum_0^k c_i A^i \vec v = \sum_0^k c_i \lambda^i \vec v = p(\lambda) \vec v. \tag{18}$
QED.
Hope this helps.  Cheers,
and as ever,
Fiat Lux!!!
A: $A v = \lambda v$ for some $v$ need not imply $A = \lambda I,$ take $A$ to be the zero matrix   $, \lambda = 1$ and $v$ any vector with two components. 
what you need to show is $A^2v = \lambda v, A^3 = \lambda^3 v, \cdots$  
to show the above string of equalities take $A v = \lambda v$ and multiply by $A$ on the left. 
A: A \vec{v} = \lambda \vec {v} doesnot imply A= \lambda.For example,
\pmatrix{2&0\0&2} \pmatrix{1\0} = 2\pmatrix{1\0}
but,
                \pmatrix{2&0\0&2} \ne 2.
Given,p(x)=c_0+c_1x+...+c_kx^k
which implies, p(A)=c_0+c_1A+...+c_kA^k
p(A)\vec v=c_0A\vec v+c_1A\vec v+...+c_kA^k\vec v \tag{1}
A \vec v = \lambda \vec v
implies  A^2 \vec v = A(A \vec v) = A(\lambda \vec v) = \lambda A \vec v = \lambda^2 \vec v
By induction,it follows that 
A^m \vec v = \lambda^m \vec v \tag{2}
From \tag{1},and \tag{2}, it follows that 
A)\vec{v}=p(\lambda)\vec{v}
Hope u understand!
