1
$\begingroup$

Let $A$ be a symmetric positive semidefinite matrix and $I$ the identity matrix.

Given the linear equation

$$ y = A(A + \sigma^2I)^{-1} x $$

Write $A$ in terms of its eigenvectors $|u_i\rangle$, $$ A=\sum_{i=1}^n \lambda_i|u_i\rangle\langle u_i| $$ and assume $$ x = \sum_{i=1}^n \gamma_i |u_i\rangle $$

How can one prove that $$ y = \sum_{i=1}^n \frac{\gamma_i\lambda_i}{\lambda_i + \sigma^2}|u_i\rangle $$

I have been trying to use the matrix inversion lemma, but I can't get the result. Is there something fundamental that I am missing?

Thanks

$\endgroup$
  • $\begingroup$ Given $A$ in the above form, and assuming the eigenvectors are normalized, $(A+\sigma^2I)^{-1}$ is $\sum\limits_{i=1}^n\frac{1}{\lambda_i+\sigma^2}\left|u_i\right>\left<u_i\right|$. $\endgroup$ – Peter Morgan Jun 8 '12 at 15:42
2
$\begingroup$

First note that \begin{align} (A+\sigma^2 I) &= \sum_i \lambda_i |u_i\rangle \langle u_i| + \sigma^2\sum_i|u_i\rangle \langle u_i|\\ &=\sum_i (\lambda_i +\sigma^2)|u_i\rangle \langle u_i| \end{align} It's diagonal with respect to $|u_i\rangle$, so the inverse is simply,

\begin{align} (A+\sigma^2 I)^{-1} &=\sum_i \frac{1}{(\lambda_i +\sigma^2)}|u_i\rangle \langle u_i| \end{align} Using the vector decomposition $x = \sum_i \gamma_i|u_i\rangle$, we have \begin{align} (A+\sigma^2 I)^{-1}x = \sum_i \frac{\gamma_i}{(\lambda_i +\sigma^2)}|u_i\rangle \end{align} and so \begin{align} A(A+\sigma^2 I)^{-1}x &=\sum_i \frac{\gamma_i}{(\lambda_i +\sigma^2)}A|u_i\rangle\\ &= \sum_i \frac{\gamma_i\lambda_i}{(\lambda_i +\sigma^2)}|u_i\rangle \end{align}

$\endgroup$
  • $\begingroup$ Gosh! First line is a saver... I did not note that! Thanks Olaf $\endgroup$ – JuanPi Jun 8 '12 at 15:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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