I am looking for a way of computing the following integral. Let $A$ be some self-adjoint complex matrix. Let $f(z) = \text{tr} \left( z I - A \right)^{-1} $. Let $\gamma$ be a simple positively-oriented closed curve that goes around some $m$ poles of $f$. The integral is:

$$ J = \oint_{\gamma} f(z) dz $$

I want to show that $J= 2 \pi i m$ somehow by pure complex analysis means without resorting to the spectral theorem, diagonalization or even eigenvalue problem. It is not true in general for the residues of a rational function.

  • $\begingroup$ What do you mean by "resorting to the resolvent formalism"? As you wrote, you're studying a resolvent. Would diagonalising it and writing the trace in terms of its eigenvalues be "resorting to the resolvent formalism"? $\endgroup$ – joriki May 15 '16 at 16:00
  • $\begingroup$ That's a misunderstanding. I don't mean "studying" in the sense of learning at a university, but "studying" in the sense of investigating, researching its properties. If you don't want this done using the eigenvalues, I think you'll need to say a bit more about how you want it done and how not. $\endgroup$ – joriki May 15 '16 at 16:08
  • $\begingroup$ Do. Not. Delete and repost. EVER! This rule is strictly upheld. Just edit this version, if you want to fine-tune or improve the question. I am merging the two versions to save the answer. $\endgroup$ – Jyrki Lahtonen May 16 '16 at 9:29
  • $\begingroup$ The rule against reposting (with or without deletion) is very strict, and it is my job to enforce the rule. The first copy stays. Always. I cannot comment about the downvote. Joriki seems to want you to clarify the meaning of resolvent formalism. I don't know if he was the downvoter, but if he was I'm sure he will reconsider if you do clarify. Don't get too worked up about a single downvote, please. $\endgroup$ – Jyrki Lahtonen May 16 '16 at 11:17
  • $\begingroup$ Oh. Too bad. I just figured out what you meant. And it's trivial with diagonalization :-( An upvote for an interesting question anyway :-) $\endgroup$ – Jyrki Lahtonen May 16 '16 at 11:26

For a piecewise smooth curve, $\gamma$, defined on $[0,1]$, the following are justified because all of the matrices involved commute: \begin{align} &\frac{d}{dt} \left((\gamma(t)I-A)\exp \left\{ -\int_{0}^{t}(\gamma(s)I-A)^{-1}\gamma'(s)ds\right\}\right) \\ &= \gamma'(t)\exp \left\{ -\int_{0}^{t}(\gamma(s)I-A)^{-1}\gamma'(s)ds\right\} \\ &-(\gamma(t)I-A)\exp\left\{-\int_{0}^{t}(\gamma(s)I-A)^{-1}\gamma'(s)ds\right\} (\gamma(t)I-A)^{-1}\gamma'(t) \\ &= 0. \end{align} Therefore, $$ (\gamma(0)I-A)=(\gamma(1)I-A)\exp\left\{-\int_{0}^{1}(\gamma(s)I-A)^{-1}\gamma'(s)ds\right\}. $$ Because $\gamma(0)=\gamma(1)$ and $(\gamma(t)I-A)$ is invertible for $0 \le t \le 1$, $$ I =\exp\left\{-\int_{\gamma}(zI-A)^{-1}dz\right\}. $$ It is well known that $\det\{\exp(A)\}=\exp\{\mbox{tr}(A)\}$. Hence, there is an integer $N$ such that $$ 1 = \det(I)=\exp\left\{-\mbox{tr}\int_{\gamma}(zI-A)^{-1}dz\right\} \\ \implies \mbox{tr}\int_{\gamma}(zI-A)^{-1}dz = 2\pi i N \\ \implies \mbox{tr}\frac{1}{2\pi i}\int_{\gamma}(zI-A)^{-1}dz = N. $$ The Functional Calculus: If $X$ is a finite-dimensional complex normed vector space, and $A$ is a linear operator on $X$, then $(\lambda I-A)^{-1}$ exists for all $\lambda\notin\sigma(A)$ and $\lim_{\lambda}\lambda(\lambda I-A)^{-1}=I$. For this reason, if $C$ is a positively-oriented simple closed rectifiable path enclosing $\sigma(A)$ in its interior, then $$ \frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1}d\lambda = I. $$ This last contour integral may be written as a sum of contour integrals around each $\lambda_k \in \sigma(A)$ because each such $\lambda_k$ is an isolated singularity of $(\lambda I-A)^{-1}$ and there are no other singularities. That's complex analysis. The integral around $\lambda_k$, say $P_k$ can be shown to be a projection, i.e., $P_k^2=P_k$. Furthermore, $P_kP_l = P_lP_k = 0$ if $k\ne l$. No diagonalization is needed to prove this. And, because of the above, $$ I = P_1 + P_2 + \cdots + P_N $$ If $X$ is an inner product space and $A$ is Hermitian, then $P_k$ is also Hermitian and you can directly show that $$ AP_k = \lambda_k P_k $$ and you can show that $Ax=\lambda_k x$ iff $P_kx=x$, just from knowing the above properties. So the range of $P_k$ is the eigenspace associated with eigenvalue $\lambda_k$. No basis is required for these facts, and no diagonalization of anything is needed. If you integrate over a simple closed positively oriented contour $C$ in $\mathbb{C}\setminus\sigma(A)$, then complex analysis gives you $$ \frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1}d\lambda = \sum_{\lambda_k \mbox{ inside } C}P_{\lambda_k} $$ So far, nothing requires a basis or diagonalization. And, by the way, you get a nice uniform operator norm approximation by integrating over contours that are allowed to enclose multiple eigenvalues. That also can be proved without basis or trace, using projections. You also get the spectral theorem because $$ I = P_{\lambda_1}+P_{\lambda_2}+\cdots+P_{\lambda_k} \\ P_{\lambda_k}^2=P_{\lambda_k}=P_{\lambda_k}^* \\ P_{\lambda_k}P_{\lambda_J}=0,\;\;\; j\ne k \\ A = \lambda_1 P_{\lambda_1}+\lambda_2 P_{\lambda_2}+\cdots+\lambda_k P_{\lambda_k}. $$

If you want to discuss trace, and you're claiming that trace requires using basis, then that's a different issue from the above. Showing $\mbox{tr}(P_{\lambda_k})$ is the rank of the projection $P_{\lambda_k}$ is not needed in the above. If that cannot be shown, then that's not an issue of calculus. What you can now show is $$ \mbox{tr}\left(\frac{1}{2\pi i}\oint_C (\lambda I-A)^{-1}d\lambda\right) = \sum_{\lambda_k\mbox{ inside } C} \mbox{tr}(P_{\lambda_{k}}) $$ But if you can use properties of trace, then you know more. However, I cannot play a game to show something about trace when you won't allow me to use basic properties of trace!!! I'll leave it at the level projections, and let you decide if you can know the trace of a projection is an integer.

  • $\begingroup$ @ValerySaharov : $(\gamma(s)I-A)^{-1}$ commutes with $A$ for all $s$. The integral $B=\int_{0}^{t}(\gamma(s)I-A)^{-1}ds$ commutes with $A$. So all powers of $B$ commute with $A$. So $e^{B}$ commutes with $A$. $\endgroup$ – DisintegratingByParts May 16 '16 at 17:31
  • $\begingroup$ @ValerySaharov : Not that I'm aware of. $\endgroup$ – DisintegratingByParts May 16 '16 at 20:59
  • $\begingroup$ @ValerySaharov : The exponential of a number is $1$ iff that number is $2\pi N$ for some integer $N$. I think you were missing the $2\pi i$. Without it, the result is not correct. $\endgroup$ – DisintegratingByParts May 17 '16 at 1:10
  • $\begingroup$ @ValerySaharov : N is the rank of the projection $P_{C}=\frac{1}{2\pi i}\oint_{C}(\lambda I-A)^{-1}d\lambda$, which is the sum of the ranks of the projections associated with $\lambda$ inside $C$. This is true for any matrix $A$, Hermitian or not. You'll notice my proof did not require Hermitian. $\endgroup$ – DisintegratingByParts May 17 '16 at 11:34
  • $\begingroup$ @ValerySaharov : You cannot diagonalize a general $A$, and the result I've given you is true for a general $A$. $P_C$ is similar to a diagonal matrix with $1$'s and $0$'s on the diagonal, even though $A$ may not be diagonalizable. $\endgroup$ – DisintegratingByParts May 17 '16 at 11:50

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