let me show you the context of my question:

Let $T:V \to V$ a linear operator, where $V$ is a finite dimensional vector space and $m\_t(x)=(x - \lambda\_1)^{m\_1} \cdots(x-\lambda\_k)^{m\_k}$.
In the proof of Primary decomposition Theorem we take $q\_j=\displaystyle\prod\_{{i=1}}^{k}(x- \lambda\_i)^{m\_i}$, with $j \neq i$ then since $gdc(q\_1,q\_2,\dots,q\_k)=1$ there exists $f\_1,f\_2,\dots,f\_k$ s.t $f\_1q\_1+f\_2q\_2+\dots +f\_kq\_k = 1$.

So, if we call $g\_j= f\_jq\_j$ we have that the operator $P\_j(T) = g\_j(T)$ is a projection s.t

1. $P\_jP\_i = 0$ if $j \neq i$
2. $I = P\_1+\cdots+P\_k$
3. $Im Pj = \ker (T - \lambda\_jId)^{m\_j}= W\_j$.

Using that we prove that $V = W\_1 \oplus\cdots \oplus W\_k$.

Ok, I have a simple question, in the condiction above, why $D = \lambda\_1P\_1 + \dots+ \lambda\_kP\_K$ is diagonalizable??

I know that for each $v \in V$, by 2. we have $v = P\_1(v)+ \cdots + P\_k(v)$ and by 1. we have $$D(v) = D(P\_1(v)+ \cdots+ P\_k(v))= (\lambda\_1P\_1(v)+ \cdots+ \lambda\_kP\_k(v))(P\_1(v)+ \cdots+ P\_k(v)) = \lambda\_1P\_1(v)+ \cdots+\lambda\_kP\_k(v).$$

I have no Idea what can I do with that, so can you help me?