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Let $V$ be a $\Bbb{K}$-vector space of dimension $n$ and $f$ be a linear operator ($f$ is diagonalizable). Let $P(x)\in \Bbb{K}[x]$ what is the characteristic polynomial of $P(f)$.

They way I did this was by showing that $f^k$ is diagonalizable for all $k\ge 1$ and that the sum of any two linear operators which are diagonalizable in the same basis is diagonalizable, then since the polynomial is a sum of diagonalizable linear operators it is also diagonalizable but I'm not really confident of the characteristic polynomial which is $$P_{P(f)}(x)=(-1)^n(x-(a_k\alpha_1^k+\ldots +a_0\alpha_1))^{r_1}\times \ldots \times (x-(a_k\alpha_q^k+\ldots +a_0\alpha_q))^{r_q}$$ where the $\alpha_i$'s are the eigenvalues , $a_i$ the coefficients of the polynomial, and $r_i$ is the multiplicity of $\alpha_i$ ($q\le n)$.

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As $f$ is diagonalizable, $V$ has a basis $v_1,\ldots, v_n$ of eigenvectors of $f$. If $f(v_k)=\lambda_k v_k$ (that is, the characteristic polynomial of $f$ is $\prod_{k=1}^n(\lambda_k-X)$), then $P(f)(v_k)=P(\lambda_k)v_k$, hence with respect to this basis $P(f)$ is a diagonal matrix with entries $P(\lambda_k)$ so that the characteristic polynomial is $$\det(P(f)-X\cdot I) =\prod_{k=1}^n (P(\lambda_k)-X).$$ If written explcitly in terms of coefficients of $P$ and multiplicities then indeed your result comes out.

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