Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.