The part of the theorem which I am not able to understand is that if $p={p_1}^{r_1}....{p_k}^{r_k}$ is the minimal polynomial of $T$.Then if $T_i$ is the operator induced on $W_i$ by $T$ then the minimal polynomial for $T_i$ is $p_i^{r_i}$.Now suppose $T$ is a diagonalizable linear operator and then $p_i$ s are all linear polynomials and let $V_1$ be the eigen space associated with $c_1$ then $E_1$ is the projection matrix on $V_1$ then it surely has the range space as $N(p_1(T))$ but how does it have the minimal polynomial as $p_1$.So does the theorem not hold for diagonalizable operators .Where am I going wrong?
Just to sum it up can some one explain the primary decomposition theorem with respect to diagonalizable operators and projection matrices $E_i$ (as described in section $6.5$,$6.7$ of Hoffman and Kunze)
I think you are mixing up the operator $T_i$ induced by $T$ on $W_i$ and the projection $E_i$ on $W_i$.
Suppose $T$ is diagonalizable. Then the minimal polynomial $p$ has the form $p=(x-c_1)...(x-c_k)$ where the $c_i$ are distinct scalars. The primary decomposition theorem says that $V=W_1 \oplus \cdots \oplus W_k$ where $W_i$ is the null space of $(T-c_iI)$. So we see that here $W_i$ is the space of characteristic vectors associated with the charateristic value $c_i$. If we take a basis $B_i$ for $W_i$, then $B=(B_1,\ldots,B_k)$ is a basis for $V$ in which the matrix of $T$ is diagonal with the $c_i$ on the diagonal entries.
Now, let $T_i$ be the operator induced by $T$ on $W_i$. By definition of $W_i$, $(T-c_iI)\beta =0$ for all $\beta \in W_i$ i.e. $T_i-c_iI_i=0$ where $I_i$ is the identity operator on $W_i$. This mean that $(x-c_i)$ annihilates $T_i$, and therefore it must be the minimal polynomial.