The representation of the resolvent of a quadratic form I know some aspects are related each other concerning resolvent ,such a system of linear equation with a parameter, Fredholm theory and Green function method in nonlinear equation when I am reading the book,Methods of Mathematical Physics Vol.1,  wrote by Courant & Hilbert. And the Spectral Theory in functional analysis is developed from the resolvent of a quadratic form originally, in some sense, according to wikipedia.
My question are focused on Page 29 of Courant's book, the resolvent of the quadratic form $K(x,x)$ is defined as 
$$ K(x,x,;\lambda) =\frac{[E(x,x) - \lambda K(x,x)]^{-1} - E(x,x)}{\lambda}$$
where $K(x,x), \; E(x,x)$are quadratic forms.
Question part 1 : What is the different between $K(x,x;\lambda)$ and $K(x,x)$ ?
Is $K(x,x;\lambda)$ a special quadratic form or a matrix? How to understand the formula from Page 18
$$E(u,y) + \lambda T(u,y;\lambda) =E(u,x)$$
Question part2:  The deduce on page 29
$$\frac{1}{\lambda}\left[ \color{red}{\left( \sum_{p=1}^n \frac{\lambda_p -\lambda}{\lambda_p} y_p^2\right)^{-1}} - E(y,y) \right] = \frac{1}{\lambda} \left[ \color{red}{\sum_{p=1}^n \frac{\lambda_p}{\lambda_p - \lambda} y_p^2} -E(y,y) \right] $$
How can the above formula be deduced from the left side to the right side? 
Thanks for help.
 A: I think I can explain that by myself.
For the matrices form, $K$ can be diagonalized, it has n numbers of eigenvectors  $\vec{l}_p$, $p=1,2, \cdots ,n$, which are orthonormalized,  with eigenvalues, $ \lambda_p$; $E$ is the unit matrices, it is easy to verify that $K$ and $E$ can be expressed in term of the eigenvectors : ( notice the sum convention)
$$ K = \frac{ \vec{l}_p \vec{l}^T_p}{\lambda_p} ;\qquad E = \vec{l}_p \vec{l}^T_p $$
so
  $$E-\lambda K =\frac{\lambda_p-\lambda}{\lambda_p}\vec{l}_p \vec{l}^T_p$$
we have the inverse matrices $$(E-\lambda K)^{-1} =\frac{\lambda_p}{\lambda_p - \lambda}\vec{l}_p \vec{l}^T_p$$
because
 $$(E-\lambda K)^{-1}(E-\lambda K) =\frac{\lambda_p}{\lambda_p - \lambda}\cdot \frac{\lambda_q-\lambda}{\lambda_q}\vec{l}_p \color{blue}{\vec{l}^T_p \vec{l}_q} \vec{l}^T_q = \frac{\lambda_p}{\lambda_p - \lambda}\cdot \frac{\lambda_q-\lambda}{\lambda_q}\vec{l}_p \color{blue}{\delta_{pq}} \vec{l}^T_q = E$$
Go back to the quadratic form:$ K \rightarrow K(x,x) = \vec{x}^T \frac{ \vec{l}_p \vec{l}^T_p}{\lambda_p} \vec{x} $ ,similar to the $E \rightarrow E(x,x)$, 
and the orthogonal transformation: $y_p= l_{qp} x_q$
or in matrices form $\vec{y}=L\vec{x}$ , so we have :
$$ K(x,x)=y^T \frac{L\vec{l}_p \vec{l}^T_p L^T}{\lambda_p} y = \sum_{p=1}^n \frac{\vec{y}^T \vec{y}}{\lambda_p}$$
therefor:
$$\big(E(x,x)-\lambda K(x,x)\big)^{-1} = \vec{x}^T \frac{\lambda_p}{\lambda_p - \lambda} \vec{l}_p \vec{l}^T_p \vec{x}= \vec{y}^T \frac{\lambda_p}{\lambda_p - \lambda} \color{green}{L\vec{l}_p \vec{l}^T_p L^T} \vec{y} =\sum_p^n \frac{\lambda_p}{\lambda_p - \lambda} y_p^2$$
notice matrics  $ L=(l_{pq})$ is an orthogonal matrix.
If you find out any improper statement above, please share your idea, thanks. 
