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Suppose that $T: V \rightarrow V$ is an endomorphism of the linear space V (about $\mathbb{K}$) and that $p(X)$ is a polynomial with coefficients in $\mathbb{K}$. Show that if $x$ is an eigenvector of $T$ than it is also an eigenvector of $p(T)$.

My attempt:

So if $x$ is an eigenvector of $T$ that means that $T(x) = \lambda x$ ($\lambda$ being the eigenvalue associated to $x$).

Ok so my next step is the one I feel is not correct

$p (T(x)) = p (\lambda x)$ so $\lambda$ is an eigenvalue of the polynomial.

I don't feel this is a correct assumption, that you can't immediately conclude this, can we?

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Regarding your attempt:
Note that the expressions $p(Tx)$ and $p(\lambda x)$ do not make sense. Indeed, except for some very special case, to the best of my knowledge, if $p$ is a polynomial $v$ is a vector, then $p(v)$ is not well defined.

Regarding the exercise:
Let $x\neq 0$ with $Tx = \lambda x$, then $$T^kx = T(T^{k-1})x=T(\lambda^{k-1}x)=\lambda^{k-1}Tx = \lambda^kx \qquad \forall k\in\Bbb N.$$

Let $p(s)=\sum_{k=0}^d \alpha_k s^k$, then $$p(T)x = \sum_{k=0}^d \alpha_kT^kx=\sum_{k=0}^d\alpha_k\lambda^kx=\Big(\sum_{k=0}^d\alpha_k\lambda^k\Big)x=p(\lambda)x.$$

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Adding a bit of generality to @Surb 's excellent accepted answer.

This exercise is just a glimpse of a very important technique in linear algebra. When $P$ is a polynomial with coefficients in $\mathbb{K}$ and $T$ is an endomorphism then $P(T)$ is also an endomorphism whose properties are closely related to those of $T$. When $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ then you can often make sense of $P(T)$ when $P$ is a convergent power series. That allows you to define useful endomorphisms like $$ e^T \quad \text{and} \quad \frac{1}{1-T}\ . $$

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Here is an outline.

Step 1: Prove by induction that if $Tx=\lambda x$, then $T^n(x)=\lambda^n x$.

Step 2: Show that if $Tx=\lambda x$ and $Sx=\mu(x)$, then $(T+S)(x)=(\lambda+\mu)x$.

Step 3: Combine the first two steps to show that if $Tx=\lambda x$, then $P(T)(x)=P(\lambda)x.$

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