Let $T$ be diagonalizable over $\mathbb{F}$. Show each projection $P_\lambda$ can be expressed as a polynomial of powers $T^j$. 
Let $T$ be diagonalizable over $\mathbb{F}$ with a spectral decomposition
  $$
T = \sum_{\lambda \in sp(T)} \lambda P_\lambda
$$
  where $P_\lambda$ is the projection onto the eigenspace $E_\lambda(T)$. Show that for each projection $P_\lambda$, there is a polynomial $f \in \mathbb{F}[x]$ such that $P_\lambda$ is a linear combination of powers $f(T) = \sum_{j=0} c_jT^j$.

Using the representation for $T$ above, we have
$$
f(T) 
= \sum_j c_j (\sum_\lambda \lambda^j P_\lambda) 
= \sum_\lambda \sum_j c_j P_\lambda^j 
= \sum_\lambda f(P_\lambda).
$$
But I don't think this gets any closer to expressing $P_\lambda$ in powers $T^j$. I'm still not even convinced that what the question is asking is possible.
 A: Let $T=\sum_{\lambda\in\sigma(T)}\lambda P_{\lambda}$, where
$$
        I=\sum_{\lambda\in\sigma(T)}P_{\lambda},\\
              P_{\lambda}^2 = P_{\lambda} \\
                  P_{\lambda}P_{\mu} = 0,\;\; \lambda\ne \mu
$$
Then,
$$
          (T-\mu I)=\sum_{\lambda\in\sigma(T)}(\lambda-\mu)P_{\lambda}
$$
If $\mu$ is equal to some $\lambda\in\sigma(T)$, then the term involving $P_{\lambda}$ does not appear on the right. Therefore, you can eliminate all but one projection in the following way:
$$
              \prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(T-\mu I)=\left(\prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(\lambda-\mu)\right)P_{\lambda} \\
       P_{\lambda} = \frac{1}{\prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(\lambda-\mu)}\prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(T-\mu I)
$$
These are the Lagrange interpolating polynomials:
$$
       P_{\lambda}=p_{\lambda}(T),\;\;\;\mbox{ where } p_{\lambda}(x) = \frac{\prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(x-\mu)}{\prod_{\mu\in\sigma(T),\;\mu\ne\lambda}(\lambda-\mu)}
$$
