I require the following $${\rm Tr}({\bf H} + \omega{\bf I})^{-1}$$ where $\bf H$ is Hermitian, $\omega$ is a complex number with a small imaginary component $(\Im \omega \approx 1\times 10^{-5})$ and $\bf I$ is the unit matrix. $\rm Tr$ indicates a trace is required

Unfortunately these matrices are $45\times 45$ and are to be numerically integrated ($\bf H$ is a function of two variables) thus I was wondering if there was a cheaper way to obtain the trace, i.e. without needing to do a full inversion.


1 Answer 1


Let $n=45$ and let $\lambda_1,...., \lambda_n$ the (real) eigenvalues of $H$. Since $H$ is diagonalisable, we have

$H=P^{-1}DP$, where $D=diag(\lambda_1,...., \lambda_n).$


$H+ \omega I=P^{-1}D_{\omega}P$, where $D_{\omega}=diag(\lambda_1+\omega,...., \lambda_n+\omega).$

Therefore $(H+ \omega I)^{-1}=P^{-1}D_{\omega}^{-1}P$, hence

${\rm Tr}((H+ \omega I)^{-1})={\rm Tr}(D_{\omega}^{-1})= \sum_{k=1}^n\frac{1}{\lambda_k+\omega}.$

  • $\begingroup$ Hi Fred, thanks for your answer, I just wanted to get some clarification: I am using the Tux Eigen package and it quotes that the computational cost to obtain the eigenvalues only, is $4n^3/3$. I can't find information as to the cost of inversion. I believe in this case it would use LU decomposition with partial pivoting. Are you able to indicate a performance gain, in $\mathcal{O} (n)$ notation? Many thanks $\endgroup$
    – AlexD
    Jan 25, 2019 at 11:32

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