# Integrating a Linear Operator $A:H\longrightarrow H$ (Matrix)

I am trying to prove a functional analysis proposition, but I got stuck. I have to integrate a matrix.

In my proof I use the following matrix:

Let $A$ be a self-adjoint matrix on $H=\mathbb{C}^n$ with the following decomposition: $$A=V^*\,\,\text{Diag}(\lambda_1,\ldots,\lambda_n)\,\,V,$$ where $V$ is Unitary and $\lambda_1\le\ldots\le\lambda_n$

Now, I came across this integral

$$f(A)=\frac{1}{2\pi i}\int_{\Gamma}{f(\lambda)(\lambda-A)^{-1}d\lambda},$$

where $\Gamma$ is the Cauchy contour around spectrum $\sigma(A)$. Here, I assume that $f$ is holomorphic on $\Omega$ (inner domain of $\Gamma$).

My question is: How can I evaluate this integral? Is there a special method to integrate matrices or an easy way to construct the contour?

Any hints are appreciated.

• I'm not sure I understand your question, so I comment. The resolvent $(\lambda I-A)^{-1}$ is a matrix rational function in $\lambda$, with poles at the eigenvalues of $A$. Thus, each component in your integral is of the form $f(\lambda)R(\lambda)$, which you can integrate with the residue theorem. Try out on a simple example: $f(\lambda)=e^\lambda$ and $$A=\begin{pmatrix} 3 & 1\\1 & 3\end{pmatrix}$$ and you will probably see how it works. – mickep Jun 13 '16 at 7:54

## 2 Answers

The contour $\Gamma$ doesn't matter, as long as all $\lambda_j$ are inside. You have, with $D=\text{Diag}\,(\lambda_1,\ldots,\lambda_n)$, $$\lambda I-A=\lambda V^* V-V^*DV=V^*(\lambda I-D)V=V^*\,\begin{bmatrix}\lambda-\lambda_1&0&\cdots&0\\ 0&\lambda-\lambda_2&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&\lambda-\lambda_n \end{bmatrix}\,V,$$ so $$(\lambda I-A)^{-1}=V^*\,\begin{bmatrix}\frac1{\lambda-\lambda_1}&0&\cdots&0\\ 0&\frac1{\lambda-\lambda_2}&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&\frac1{\lambda-\lambda_n} \end{bmatrix}\,V.$$ Conjugation by $V$ is linear and continuous, so \begin{align} f(A)&=\frac1{2\pi i}\,\int_\Gamma f(\lambda)\,V^*(\lambda I-D)^{-1}\,V\,d\lambda =V^*\,\frac1{2\pi i}\,\int_\Gamma f(\lambda)\,(\lambda I-D)^{-1}\,d\lambda\,V\\ \ \\ &=V^*\,\begin{bmatrix}\displaystyle\frac1{2\pi i}\,\int_\Gamma \frac{f(\lambda)}{\lambda-\lambda_1}\,d\lambda&0&\cdots&0\\ 0&\displaystyle\frac1{2\pi i}\,\int_\Gamma \frac{f(\lambda)}{\lambda-\lambda_2}\,d\lambda&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&\displaystyle\frac1{2\pi i}\,\int_\Gamma \frac{f(\lambda)}{\lambda-\lambda_n}\,d\lambda \end{bmatrix} \, V\\ \ \\ &=V^*\,\begin{bmatrix}f(\lambda_1)&0&\cdots&0\\ 0&f(\lambda_2)&\cdots&0\\ \vdots& &\ddots&\vdots \\0&0&\cdots&f(\lambda_n) \end{bmatrix} \, V \end{align}

• That is not a scoop (cf. my answer) and that is not the question. How do you choose a contour if, concerning the eigenvalues, you know only some bounds (using, for example, the Gerschgorin's discs) ? – loup blanc Jun 13 '16 at 22:31
• Not totally sure what you mean. For this to make sense, as stated in the question, you need $\Gamma$ to a curve that contains the whole spectrum $\sigma(A)$. This is not that hard, since the spectrum consists of $n$ points. The assumption in the question is that $f$ is holomorphic in the interior of $\gamma$. – Martin Argerami Jun 13 '16 at 22:35
• Yes I understand your point of view. In fact, I think that the OP did not know the formula $f(A)=V^*diag(f(\lambda_i))V$!! What interests me is the numerical calculation that could be done by setting up an ad hoc contour. Moreover the OP's condition does not suffice. One must also have (at least) the continuity of $f$ over $\overline{\Omega}$. – loup blanc Jun 13 '16 at 23:40
• I cannot say anything about numerical computations. The continuity on the boundary is not a big deal, at least from a theoretical point of view: here $\sigma(A)$ is discrete, so if $f$ is holomorphic on the interior, one can always shrink the curve a little. – Martin Argerami Jun 14 '16 at 0:09

Case 1. You know approximations of the eigenvalues of $A$; then you know also (the calculation is in $O(n^3)$) a convenient matrix $V$ and finally $f(A)=V^*diag(f(\lambda_1),\cdots,f(\lambda_n))V$.

Of course, you can also use the mickep's method. Yet, the calculation of $(\lambda I_n-A)^{-1}$ becomes complicated when $n$ grows.

Case 2. You know only bounds about the eigenvalues; then you choose some disc containing $spectrum(A)$ and no poles of your function $f$. Parametrize the edge of the disc $\lambda=a+re^{i\theta},\theta\in[0,2\pi]$. Calculate (numerically) $1/(2i\pi)\int_0^{2\pi}f(a+re^{i\theta})((a+re^{i\theta})I_n-A)^{-1}rie^{i\theta}d\theta$.