Why is $\mathbf{C}=(\lambda\mathbf{I}-\mathbf{A})^+(\lambda\mathbf{I}-\mathbf{A})+\mathbf{x}\mathbf{x}^\top$ symmetric idempotent? Let $\mathbf A\mathbf x=\lambda\mathbf x$ be the eigenvalue equation for a real symmetric $\mathbf A \in\mathbb R^{n\times n}$. We now define $\mathbf B=(\lambda\mathbf I-\mathbf A)$ and $\mathbf C=\mathbf B^+\mathbf B +\mathbf x\mathbf x^\top$, where $\mathbf B^+$ is the Moore-Penrose pseudoinverse. In addition, $\mathbf x^\top\mathbf{x}=1$ holds true. Show that $\mathbf C$ is symmetric and idempotent.
I am really stuck on this one. Any help is appreciated.
 A: Assuming that $x$ is a the normalized eigenvector it follows:
\begin{align}
  C^2 = B^+ \underbrace{B B^+ B}_{=B}
      + B^+ \underbrace{B x}_{= 0} x^T
      + \underbrace{xx^T B^+ B}_{= xx^T B^T (B^+)^T = 0}
      + x\underbrace{x^T x}_{= \|x\|^2 = 1} x^T
      = B^+ B + xx^T
      = C
\end{align}
Addendum: I found a simple proof for JeanMarie's conjecture: If $A$ is symmetric, invertible and has no double eigenvalues, then $C=I$.
Since $A$ is real symmetric there exists an orthogonal matrix $P$ with $PAP^{-1} = D$, where $D=\text{diag}(\lambda_1,\ldots,\lambda_n)$ is the diagonal matrix containing the eigenvalues of $A$.
Now without loss of generality let $\lambda = \lambda_1$ be the first eigenvalue of $A$ and $x$ the appropriate normalised eigenvector. Moreover note that
$$ (\lambda I - A)^+ = \big(P^{-1}(\lambda I - D) P\big)^+ = P^{-1}(\lambda I - D)^+ P $$
Thus
\begin{align}
C = B^+B + xx^T &=\big(P^{-1}(\lambda I - D)^+(\lambda I - D) P + xx^T\big) \\
&= P^{-1}\big((\lambda I - D)^+(\lambda I - D)  + (Px)(Px)^T\big)P \\
&= P^{-1}P = I
\end{align}
Because
$$(\lambda I - D)^+(\lambda I - D) = \small{
\begin{pmatrix} 0 &&& \\&1/\lambda_2&&\\&&\ddots&\\&&&1/\lambda_n\end{pmatrix}
\begin{pmatrix} 0 &&& \\&\lambda_2&&\\&&\ddots&\\&&&\lambda_n\end{pmatrix}
=
\begin{pmatrix} 0 &&& \\&1&&\\&&\ddots&\\&&&1\end{pmatrix}} $$
And 
$$(Px)(Px)^T = e_1 e_1^T = \begin{pmatrix} 1 &&& \\&0&&\\&&\ddots&\\&&&0\end{pmatrix} $$
Since $Ax=\lambda_1 x \iff PAP^{-1} P x = \lambda_1 P x \iff DPx = \lambda_1 P x \implies Px = \pm e_1$
A: In fact, in a large majority of cases (see our exchange with @Hyperplane) ** $C $ the identity matrix.** Thus, in particular it is symmetrical and idempotent.
This is attested by many simulations using random matrices of different orders with the following Matlab program:

clear all;close all
n=3;A=rand(n);
A=A'+A; % made symmetrical
[P,D]=eig(A);L=D(1,1); % first eigenvalue 
x=P(:,1); % associated (normalized) eigenvector
B=L*eye(n)-A; 
C=pinv(B)*B+x*x'


I confess that I have at present no proof of this property..
