A question on the spectral projection I am reading a paper about spectral theory. And I meet with some problems.
An operator $K\in L(X)$ is said to be algebraic if there exists a non-trivial complex polynomial $h$ such that $h(K)=0$. By the spectral mapping theorem, $h(\sigma(K))=\sigma(h(K))=\{0\}$, so that $\sigma(K)$ is finite, say $\sigma(K)=\{\mu_{1},...,\mu_{n}\}$. For $i=1,..., n$, let $P_{i}\in L(X)$ denote the spectral projection associated with $K$ and with the spectral set $\{\mu_{i}\}$, and let $Y_{i}:=R(P_{i})$. 
From standard spectral theory it is known that $P_{1}+...+P_{n}=I$, that $Y_{1},...,Y_{n}$ are closed linear subspaces of $X$ which are each invariant under $K$, and that $X=Y_{1}\oplus...\oplus Y_{n}$.(I do not know why.)
My question is: Are the properties above the classical theorems in spectral theory? In which book we can find these conclusions?
 A: The existence of such a polynomial $h$ is what gives the result that you want. For this discussion, assume $X$ is a complex Banach Space. The projections are not projections onto eigenspaces, but they're Jordan-type projections. More specifically, if the minimal annihilating polynomial for $K$ is
$$
    h(\lambda) = (\lambda-\lambda_{1})^{r_{1}}\cdots(\lambda-\lambda_{n})^{r_{n}},
$$
then there are projections $P_{j}$ (i.e., $P_{j}^{2}=P_{j}$) which are polynomials in $K$
 (and, hence, commute with $K$) such that
$$
         P_{j}P_{k}=P_{k}P_{j} = 0 \mbox{ for } j \ne k.
$$
$$
             P_{1}+P_{2}+\cdots+P_{n} = I.
$$
$$
             (A-\lambda_{j}I) \mbox{ is nilpotent on } P_{j}X \mbox{ of order } r_{j}.
$$
The last statement is equivalent to saying that $(A-\lambda_{j}I)^{r_{j}}P_{j}=0$ and $(A-\lambda_{j}I)^{r_{j}-1}P_{j} \ne 0$. To see that these projections exist, notice that the polynomials
$$
          m_{k}(\lambda) = \prod_{j\ne k}(\lambda-\lambda_{j})^{r_{j}}
$$
have no common factors and, therefore, there exist polynomials $p_{k}$ such that
$$
               p_{1}m_{1}+\cdots +p_{n}m_{n} = 1.
$$
Let $P_{j}=p_{j}(K)m_{j}(K)$. It follows that $\sum_{j=1}^{n}P_{j}=I$ and $P_{j}P_{k}=P_{k}P_{j}=0$ for $j\ne k$. It's also clear that $(A-\lambda_{j})^{r_{j}}P_{j}=0$. In fact, $P_{j}$ is a projection, i.e., $P_{j}^{2}=P_{j}$. Furthermore, if $P=p(K)$ is a projection for some polynomial $p$, then, for any $j$, either $PP_{j}=P_{j}$ or $PP_{j}=0$.
You should be able to verify these facts for yourself, but I don't have a good reference for you.
