Reverse spectral theorem proof 
Let $T$ be a linear transformation in a finite inner product space $V$ , let $l_1,...l_k$ be different scalars , and let $P_1 \not=0,....,P_k \not=0$ linear transformations in $V$ that satisfies the following conditions:

*

*$T= l_1P_1+...+l_kP_k$

*$I=P_1+...+P_k$

*for all $i \not = j$ $P_i P_j =0$

*for all $i$ $P_i^2=P_i$

*for all $i$ $P_i^*=P_i$.

the first condition in $T$ is called "The spectral theorem of $T$ ".
let $T$ be a linear transformation in $T$ in a finite inner product space $V$ , and let $l_1P_1+...+l_kP_k$ be a spectral decomposition of $T$. prove that: $\{l_1,...,l_k\}$ is the set of all eigenvalues of $T$; $P_1,...,P_k$ are orthographic projection to the eigen
subspace that belong to $l_1,...,l_k$ accordingly ; $T$ is normal when $F= \Bbb C$ and $T$ is symmetric when $F= \Bbb R$


I saw this post which is the same as mine but I am still confused a bit because of the solution provided by my professor, this is his solution:
First we will prove that $P_i$ is an orthonormal projection, let $V_i = \{v | P_i v =v \}$ and $W_i = \{v | P_i v =0 \}$  then $V_i \cap W_i = \{0\}$ (this is my first question , why is that? what do $W_i$ and $V_i$ represent here? why did he define them this way?) . for all $v \in V$ : $v= P_i v +(v- P_i v)$.
since $P_i^2 =P$ then $P_i(P_iv)=P_iv$ therefore $ P_iv \in V_i$ (why is $ P_iv \in V_i$?)  also $P_i (v-P_iv)=0$ so $(v-P_iv) \in W_i$ (I guess this is because $T(v)=0 \iff v=0$?)
from our conclusions here we get $V= V_i \oplus W_i$.
let $w \in W_i$ be any vector and let $v \in V_i$ then
$(v,w)=(P_iv,w)=(v,P_i^*w)=(v,P_iw)=(v,0)=0 $ (is this because of the standard inner product and the last condition? (#5)) , therefore because of this and dimension consideration we get $W_i = V_i^\perp$ finally we get that $P_i$ is an orthogonal projection on $V_i$.
for all $j \not = i$ and for all $v \in V_j$ we get $P_iv = P_u P_jv = 0(v)=0$ (third condition)  so $V_j \subseteq W_i$ and $V_i \perp V_j$ and $V_i \cap V_j = \{0\}$.
let $v \in V$ then $v=Iv= (P_1+...+P_k)v = P_1v +...+ P_kv$ (second condition) and from here we get $V+ V_1+V_2+...+V_k$ from $V_i \cap V_j = \{0\}$ and $V+ V_1+V_2+...+V_k$ we conclude that $V= V_1 \oplus ... \oplus V_k$
according to the first condition for all $0 \not = v \in V_i$ we get $Tv= (l_1P_1+...+l_kP_k)v=l_iP_iv=l_iv$ because $P_jv=0$ for $j \not i$ ( there exists a vector $v \not = 0 $ because $P_i \not = 0$) therefore $v$ is an eigenvector of $T$ that belongs to the eigenvalue $l_i$ , if we define the space $U_i$ of the eigenvalue $l_i$ then we get $V_i \subseteq U_i$ , also let $v \in U_i$.
$V= V_1 \oplus ... \oplus V_k$ so we can define $v=v_1+...+v_k$ , $v_t \in V_t$ for $1 \leq t \leq k$. so $Tv= l_iv=...=l_iv_k$ but $Tv=l_1 v_1 +...+ l_k v_k$ so $(l_i-l_1)v_1 +...+ (l_i - l_k)v_k=0$ and because of the uniqueness of a direct sum we conclude that for all $j$ $(l_i - l_j) v_j =0$ if $j \not = i$ then $l_j \not = l_i$ so we conclude that $v_j =0$ and $v=v_i$ meaning $v \in V_i$. that way we showed that $U_i \subseteq V_i$ and $V_i = U_i$.
lastly given for $i=1,...,k$ $T^*= (\sum l_i P_i)^* = \sum \bar l_i P_i^* = \sum \bar l_i P_i^*$ if we are in the real number field we get $\bar l_i = l_i$ and then $T^*=T$ so $T$ is symmetric.
and in complex field $TT^*=T^*T=\sum|l_i|^2 P_i$ because $P_i^2=P_i$ and $P_iP_j=0$ so it is normal.
Hopefully my translations of his solution are correct.. I am aware that his solution is detailed but appreciate any further explanations for to make it simpler and more understandable or any other (not so advanced please) way is acceptable. hopefully his solution can help someone else on the forum though.
 A: You have a lot of questions here, but your comment seems to indicate you are mainly interested in the first two, so I'll answer those in the meantime. Feel free to let me know if you have follow-up questions and I can edit this to address those as well.

*

*The idea behind defining $V_i$ and $W_i$ as they are is that condition (4) implies that the eigenvalues of each $P_i$, should they exist, are either $0$ or $1$ (do you see why?). As such, we are interested in the set of vectors that are fixed by $P_i$ (corresponding to eigenvalue $1$; this is $V_i$), and the set of vectors that are zeroed out by $P_i$ (corresponding to eigenvalue $0$; this is $W_i$).


*Since $P_i^2=P_i$ for all $v$, we know that $P_i(P_i v)=P_i v$, and as such $P_i v$ is fixed by $P_i$, which is precisely the definition of being in $V_i$. Similarly, $P_i(v-P_i(v))=P_i v-P_i^2 v = P_iv-P_iv=0$, so $v-P_iv$ is zeroed out by $P_i$, which is precisely the definition of being in $W_i$. The upshot of this analysis is that we now know any vector in $V$ can be written as the sum of a vector in $V_i$ and a vector in $W_i$.
Hope this helps!
EDIT: For your next question:


*Yes, exactly. We have $(v,w)=(P_iv,w)$ because $P_iv=v$ by the definition of $V_i$, $(P_iv,w)=(v,P_i^*w)$ by the definition of the adjoint, $(v,P_i^*w)=(v,P_i w)$ by condition (5), and $(v,P_i w)=(v,0)=0$ because $P_iw=0$ by the definition of $W_i$.

EDIT 2:


*I'm going to assume that you understand the proof that $V=V_1\oplus\dots\oplus V_k$, as it is very similar to the work before involving $V_i$ and $W_i$. The thing to note is that since all of the $V_i$'s are orthogonal and each $P_i$ is an orthogonal projection, if we pick $v\in V_i$ for some $i$, all of the terms in $Tv$ are going to be zero except for $l_iP_iv$, which will equal $l_i v$ since $P_iv=v$. Therefore $v$ is an eigenvector of $T$ with corresponding eigenvalue $l_i$. Doing this for each $V_i$, we construct a list of $k$ linearly independent eigenvectors with corresponding eigenvalues $l_1,\dots,l_k$ and eigenspaces that span $V$. Since the eigenspaces span $V$, we know we have found all possible eigenvalues.

