Functions of a diagonalizable matrix

(1) Polynomial case. For $f(X)=\sum_ka_kX^k$, using $A^k=PD^kP^{-1}$:
$$f(A)=\sum_ka_kPD^kP^{-1}=P\Big(\sum_ka_kD^k\Big)P^{-1}=P\,f(D)\,P^{-1},$$
since $\sum_ka_kD^k=\textrm{diag}\big(f(\lambda_j)\big)$.

(2) Well-definedness. Every diagonal entry $\lambda_j$ of $D$ is one of the $\mu_i$. Let $q$ be a polynomial with $q(\mu_i)=f(\mu_i)$ for $i=1,\ldots,r$ (e.g. $q=\sum_if(\mu_i)\prod_{j\neq i}\frac{X-\mu_j}{\mu_i-\mu_j}$). Then $f(D)=q(D)$, and by (1)
$$P\,f(D)\,P^{-1}=P\,q(D)\,P^{-1}=q(PDP^{-1})=q(A).$$
The right-hand side depends only on $A$ (and $f$ on the spectrum), not on the chosen $P$ and $D$.

(3) Example. $A(1,-1)^T=(1,-1)^T$ and $A(1,1)^T=3(1,1)^T$, so $A=PDP^{-1}$ with $P=\begin{pmatrix}1&1\\-1&1\end{pmatrix}$, $D=\textrm{diag}(1,3)$, $P^{-1}=\frac12\begin{pmatrix}1&-1\\1&1\end{pmatrix}$. Then
$$\sqrt A=P\textrm{diag}(1,\sqrt3)P^{-1}=\frac12\begin{pmatrix}1+\sqrt3&\sqrt3-1\\\sqrt3-1&1+\sqrt3\end{pmatrix}.$$
Check: the $(1,1)$ entry of $(\sqrt A)^2$ is $\frac14\big((1+\sqrt3)^2+(\sqrt3-1)^2\big)=\frac{8}{4}=2$ and the $(1,2)$ entry is $\frac14\cdot2(1+\sqrt3)(\sqrt3-1)=1$, as required.

Conclusion. Functions of a diagonalizable matrix are well defined through its eigen-decomposition and coincide with polynomial functions of $A$ on the spectrum. For a power series such as $e^{tA}=\sum_kt^kA^k/k!$ one gets $e^{tA}=P\textrm{diag}(e^{t\lambda_j})P^{-1}$. For non-diagonalizable matrices one uses the Jordan form instead.