Show that the diagonal entries of symmetric & idempotent matrix must be in [$0,1$] Show that the diagonal entries of symmetric & idempotent  matrix must be in [$0,1$].
Let $A$ be a symmetric and idempotent $n \times n$ matrix. By the definition of eigenvectors and since $A$ is an idempotent,
$Ax=\lambda x \implies A^2x=\lambda Ax \implies Ax=\lambda Ax=\lambda^2 x.$
So $\lambda^2=\lambda$ and hence $\lambda \in \{0,1\}$. To show the part about the "diagonal matrix" I use the fact that every symmetric matrix is diagonalizable.
Is this a complete proof?
 A: Expanding my comment to an answer, as OP appears to have lost interest: 
Recall the hypotheses: $A$ is $n\times n$, idempotent (so $A^2=A$), and symmetric (so $a_{ij}=a_{ji}$, if we let $a_{ij}$ be the entry in row $i$, column $j$ of $A$). 
Looking at the entry in row $i$, column $i$ on both sides of $A=A^2$ we get $$a_{ii}=a_{i1}^2+a_{i2}^2+\cdots+a_{ii}^2+\cdots+a_{in}^2\ge a_{ii}^2$$ But the inequality $a_{ii}\ge a_{ii}^2$ is equivalent to $0\le a_{ii}\le1$. 
A: Let $Q$ be a real symmetric and idempotent matrix of "dimension" $n \times n$. First, we establish the following:
The eigenvalues of $Q$ are either $0$ or $1$.
proof. note that if $(\lambda,v)$is an eigenvalue- eigenvector pair of $Q$ we have 
$\lambda v=Qv= Q^{2} v=Q(Qv)=Q(\lambda v) = \lambda^{2} v$.  Since $v$ is nonzero then the result follows immediately.
With this result at hand the following observation gets us to the desired answer:
Let $e_{i}$ and $q_{ii}$ denote the standard unit vector and $i_{th}$ diagonal element of $Q$, respectively. Then we have 
$$
0= min \{\lambda_1 ,...,\lambda_{n} \} \leq q_{ii} = e_{i}' Q e_{i} \leq max \{\lambda_{1},...,\lambda_n\}=1
$$
A: All you have shown till now (using idempotency) is that the eigenvalues are either $0$ or $1$.
Now we use symmetry to say that your matrix (let us call it $A$) is unitarily similar to the diagonal matrix consisting of eigenvalues of $A$ on its diagonal. That is
$$A = UDU^{-1}$$
for some unitary matrix $U$ and the diagonal matrix $D$ where $D$ has eigenvalues of $A$ on its diagonal. We already know that these eigenvalues are either $0$ or $1$. So now expand the above representation of $A$ to get the diagonal entries of $A$.
