TITLE: Understanding Spectral Theorem
QUESTION [10 upvotes]: I got the proof from mentioned here.

Spectral Theorem: Let $T$ be a normal operator on a finite-dimensional complex inner product space $V$ or a self-adjoint operator on a finite-dimensional real inner product space $V$.
  Let $c_1, \dotsc, c_k$ be the distinct characteristic values of $T$.
  Let $W_j$ be the characteristic space associated with $c_j$ and $E_j$ the orthogonal projection of $V$ on $W_j$.
  Then $W_j$ is orthogonal to $W_i$ when $i \neq j$, $V$ is the direct sum of $W_1, \dotsc, W_k$, and
  $$
  T = c_1 E_1 + \dotsb + c_k E_k$$

Proof: Let $\alpha$ be a vector in $W_j$, $\beta$ a vector in $W_i$, and suppose $i\neq j$. Then $c_j⟨\alpha,\beta⟩=⟨T\alpha,\beta⟩=⟨\alpha,T^*\beta⟩=⟨\alpha,\overline{c_j}\beta⟩$. Hence $(c_j-c_i)⟨\alpha,\beta⟩=0$, and since $c_j-c_i\neq 0$, it follows that $⟨\alpha,\beta⟩=0$. Thus $W_j$ is orthogonal to $W_i$ when when $i\neq j$. $\color{red}{\text{From the fact that } V \text{ has an orthonormal basis consisting of characteristic vectors, it}}$ $\color{red}{\text{follows that }V=W_1+\cdots+W_k}$. If $\alpha_j$ belongs to $V_j(1\leq j\leq k)$ and $\color{red}{\alpha_1+\cdots+\alpha_k\stackrel{?}{=}0}$, then
\begin{align}
0&\color{red}{\stackrel{?}{=}}(\alpha_i,\sum_j\alpha_j)=\sum_j(\alpha_i,\alpha_j)\qquad(1)\\
&=\|\alpha_i\|^2
\end{align}
For every $i$, so that $V$ is the direct sum of $W_1,\cdots,W_k$. Therefore $\color{red}{E_1+\cdots+E_k=I}$ and 
\begin{align}
T&=TE_1+\cdots+TE_k\\
&=c_1E_1+\cdots+c_kE_k
\end{align}
I am not a follower of Hoffman and Kunze's linear algebra but for seeking my proof of Spectral Theorem I read that part. And I didn't manage to understand the red marked line. Like:
$(1)$ Why $V$ space can be written as a sum of the $W_i$ space$?$
$(2)$ Why the sum of the vectors $\alpha_i$ is zero$?$
$(3)$ Why the equality hold for $(1)?$
$(4)$ Why the sum of eigenvectors $E_i$ is $I?$
Maybe all of my question require another reference theorem but I really want to know all of those Why.
Thanks for your time and thanks in advance.

REPLY [3 votes]: Here's an example:
$$ P = 
\begin{pmatrix}
0 & \frac1{\sqrt2} &  \frac1{\sqrt2} \\
0 & \frac1{\sqrt2} & -\frac1{\sqrt2} \\
1 & 0              & 0 
\end{pmatrix};\;
T = P
\begin{pmatrix}
-1 & 0 & 0 \\
 0 & -1 & 0 \\
 0 & 0 & 2
\end{pmatrix}
P^T =
\begin{pmatrix}
\frac12 & -\frac32 & 0 \\
-\frac32 & \frac12 & 0 \\
0 & 0 & -1
\end{pmatrix}.
$$
There are two eigenvalues: $c_1 = -1$ and $c_2 = 2$. The orthogonal projections are
\begin{align}
E_1 &= P
\begin{pmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 0
\end{pmatrix}P^T =
\begin{pmatrix}
\frac12 & \frac12 & 0 \\
\frac12 & \frac12 & 0 \\
0 & 0 & 1
\end{pmatrix}, \\
E_2 &= P
\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1
\end{pmatrix}P^T =
\begin{pmatrix}
\frac12 & -\frac12 & 0 \\
-\frac12 & \frac12 & 0 \\
0 & 0 & 0
\end{pmatrix}.
\end{align}
This works in general. To obtain $E_i$, you take $T = PDP^T$ and replace all the eigenvalues in $D$ by $1$ if they equal $c_i$ and $0$ if they are a different eigenvalue.
You should convince yourself that

$$ E_1 + E_2 + \dots + E_k = P
\begin{pmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1
\end{pmatrix} P^T = I.
$$
$$ c_1 E_1 + c_2 E_2 + \dots + c_k E_k = PDP^T = T. $$
$E_i$ projects onto the $c_i$-eigenspace. For a start: if you multiply $P$ times the modified diagonal matrix, the remaining columns of $P$ are a basis for the $c_i$-eigenspace, in particular the column space (image) of $E_i$ is contained in the $c_i$-eigenspace.

Comments:

This is all easy to see if you just don't write the $P$'s and $P^T$'s and just write the matrices in terms of the eigenbasis rather than the standard basis.
This works for any diagonalizable matrix, if you replace $P^T$ by $P^{-1}$. I'm just using a self-adjoint (i.e. symmetric) operator because that's what the Spectral Theorem applies to. For self-adjoint matrices the special facts are: 1) they are always diagonalizable 2) we can diagonalize them with an orthogonal matrix.