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?


  • $\begingroup$ 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

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}

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

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