Let $T$ be a linear operator on a finite dimensional complex vector space $V$. If
1) If $p(x)=a_nx^n+...+a_1x+a_0$, then $p(T):=a_nT^n+...+a_1T+a_0I$.
2) If $f(x)=\sum_{n=0}^{\infty}a_nx^n$ is analytic, then $f(T):=\sum_{n=0}^{\infty}a_nT^n$.
Now suppose $T$ is normal. Then $V=\oplus_{i=1}^n E_i$ and $f=\lambda_1\rho_1+...+\lambda_n\rho_n$ where $\lambda_i$ is an eigenvalue of $T$ and $\rho_i$ is a projection onto $E_i$. If $f$ is an arbitrary function (defined on all of $\mathbb{C}$) define $f(T)=f(\lambda_1)\rho_1+...+f(\lambda_n)\rho_n$. Can we show that this definition is equivalent to 1) and 2) above when $f$ is a polynomial/analytic?
Let $f(x)=x^k$ , $A:=\lambda_1^k\rho_1+...+\lambda_n^k\rho_n$ and $B:=T^k$.
It suffices to show that $A=B$ (why ?)
If $x \in V$, then $x=u_1+...+u_n$, with $u_j \in E_j$.
$Bx=\lambda_1^ku_1+...+\lambda_n^ku_n$.
Since $\rho_ju_{\mu}=\delta_{j \mu}u_j$,
$Ax=\lambda_1^ku_1+...+\lambda_n^ku_n$.
Therefore $Ax=Bx$ for all $x \in V$