Reference Request: Good Introduction to Functional Calculus I would like to know if there is a standard reference on functional calculus. In particular I am interested in understanding how the resolvant integral
$$f(T) = \frac{1}{2\pi i}\int_{C} f(\lambda)[\lambda - T]^{-1}d\lambda$$
is used to derive formulas like the polynomial formula for $N\times N$ diagonalizable operators, 
$$P(T) = \sum_{j=1}^NP(\lambda_j)\prod_{k\neq j}\frac{\lambda_k-T}{\lambda_k-\lambda_j}$$
and in particular to calculate arbitrary powers of $N\times N$ diagonalizable matrices. 
 A: If $T$ is a diagonalizable matrix, then the minimal polynomial $m$ for $T$ has no repeated roots. Because of this, the resolvent $(T-\lambda I)^{-1}$ has simple poles at the eigenvalues. To see why, note that
$$
         m(\mu)-m(\lambda)=(\mu-\lambda)q(\lambda,\mu),
$$
where $q(\lambda,\mu)=q(\mu,\lambda)$ and $q(\lambda,\mu)$ is a polynomial in $\mu$ for fixed $\lambda$. Consequently, evaluating at $\mu=T$ gives
$$
            m(T)-m(\lambda)I = (T-\lambda I)q(\lambda,T) \\
              -m(\lambda)I = (T-\lambda I)q(\lambda,T)
$$
Hence, $(T-\lambda I)$ is invertible iff $m(\lambda)\ne 0$ and, in that case,
$$
        (T-\lambda I)^{-1}= -\frac{1}{m(\lambda)}q(\lambda,T),
$$
the right side of which has first order poles at the eigenvalues.
Therefore, if $f$ is holomorphic on an open region containing the eigenvalues of $T$, and if $C$ is a positively oriented system of contours enclosing the eigenvalues, then
$$
     f(T) = \frac{1}{2\pi i}\oint_{C}f(\lambda)(\lambda I-T)^{-1}d\lambda
$$
is determined by the values of $f$ at the eigenvalues only. In particular, if $p_j$ is a polynomial that is $1$ at an eigenvalue $\lambda_j$ and is $0$ at all other eigenvalues, then
$$
                \sum_{j}p_j(T)=I,\;\;\; p_j(T)^2=p_j(T), \\
                p_j(T)p_k(T)=0 \mbox{ for } j\ne k.
$$
For example, if $\{ \lambda_1,\lambda_2,\cdots,\lambda_n \}$ are the eigenvalues of $T$, then examples of $p_j$ are the Lagrange interpolating polynomials
$$
            p_j(z) = \frac{\prod_{k\ne j}(z-\lambda_k)}{\prod_{k\ne j}(\lambda_j-\lambda_k)}, \;\;\; p_j(\lambda_k) = \delta_{j,k}.
$$
Therefore,
$$
                  p_j(T)=\frac{1}{\prod_{k\ne j}(\lambda_j-\lambda_k)}\prod_{\ne k}(T-\lambda_kI)
$$
If $f$ is an arbitrary holomorphic function as described above, then
$$
              g(z)=f(z)-\sum_{j=1}^{n}f(z_j)p_j(z)
$$
satisfies $g(T)=0$. Therefore,
$$
               g(T) = f(z_j)p_j(T),
$$
which is what you wanted to show.
Reference: Taylor and Lay, Introduction to Functional Analysis
