The spectral mapping theorem for polynomials

Triangularisation (Schur). Over $\mathbb{C}$ there is an invertible $P$ with $A=PTP^{-1}$, $T$ upper triangular with diagonal entries $\lambda_1,\dots,\lambda_n$, the eigenvalues of $A$ repeated according to multiplicity.

Powers. $A^k=PT^kP^{-1}$ for all $k\ge0$. A product of upper triangular matrices is upper triangular, with diagonal entries equal to the products of the diagonal entries; so $T^k$ is upper triangular with diagonal $(\lambda_i^k)$.

Polynomials. If $p(x)=\sum_kc_kx^k$, then $p(A)=\sum c_kPT^kP^{-1}=P\,p(T)\,P^{-1}$, and $p(T)=\sum c_kT^k$ is upper triangular with diagonal entries $\sum c_k\lambda_i^k=p(\lambda_i)$.

Conclusion. The characteristic polynomial of $p(A)$ is that of $p(T)$, namely $\prod_i(t-p(\lambda_i))$. Therefore the spectrum of $p(A)$ is $\{p(\lambda_1),\dots,p(\lambda_n)\}=p(\operatorname{spec}A)$.

Example. If $A$ is nilpotent, $\operatorname{spec}A=\{0\}$, so $A^2+I$ has spectrum $\{1\}$ and is invertible. If $A^2=A$ (projection), $\operatorname{spec}A\subset\{0,1\}$ because $p(x)=x^2-x$ gives $p(\operatorname{spec}A)=\operatorname{spec}(0)=\{0\}$.

Remark. The real spectrum is not enough: $A=\begin{pmatrix}0&-1\\1&0\end{pmatrix}$ has no real eigenvalue but $p(x)=x^2+1$ gives $p(A)=0$. The theorem must be applied over $\mathbb{C}$.