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Let $A$ and $B$ be two real $n\times n$ matrices, and let $C(x)=B(I-xB)^{-1}$, where $I$ is the identity matrix of order $n$, for any real scalar $x$ such that $I-xB$ is invertible. Denote by $\mathrm{tr}$ the trace operator.

Is it possible to get a closed form solution for

$$\int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx},$$

at least when $B$ is diagonalizable?


What I've done so far:

Assuming $B$ is diagonalizable, $C$ is too, and hence it admits the spectral decomposition

$$ C(x)=\sum_{\lambda\in\mathrm{Sp}(B)}\frac{\lambda}{1-x\lambda}Q_{\lambda}, $$

where $\mathrm{Sp}(B)$ denotes the set of distinct eigenvalues of $B$, and $Q_{\lambda}$ is the projector onto $\mathrm{null}(B-\lambda I)$ along $\mathrm{col}(B-\lambda I)$ ($\mathrm{null}$ and $\mathrm{col}$ stand for the null and the column spaces). Hence

$$ \int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}=\sum_{\lambda\in \mathrm{Sp}(B)}\mathrm{tr}(AQ_{\lambda})\int\frac{\lambda}{1-x\lambda}\,dx $$

If all the eigenvalues of $B$ are real we obtain

$$ \int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}=\sum_{\lambda\in \mathrm{Sp}(B)}\mathrm{tr}(AQ_{\lambda})\int\frac{\lambda}{1-x\lambda }\,dx=\sum_{\lambda\in\mathrm{Sp}(B)}\mathrm{ln}(\left\vert 1-x\lambda \right\vert )\mathrm{tr}(AQ_{\lambda}) $$

But what about the case in which not all eigenvalues of $B$ are real? Do we get any simplification from the fact that the eigenvalues of real matrices come in complex conjugate pairs, and the eigenvectors (and hence the projectors $Q_{\lambda}$) associated to complex conjugate eigenvalues are complex conjugates? Can we still use the formula in the last display above for $\int{\mathrm{tr}}\left( {AC(x)}\right) {\,dx}$?

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  • $\begingroup$ Why is there an $x$ remaining after you have worked out the integral? It is meant to be a definite integral, right? But then what is the domain of integration? $\endgroup$ – Justpassingby Jan 14 '16 at 21:42
  • $\begingroup$ It is meant to be an indefinite integral. Is there anything unclear in the question? $\endgroup$ – mark Jan 15 '16 at 0:04
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Let $f:x\in I\subset \mathbb{R}\rightarrow -tr(A\log(I-xB)),g(x)=\log(I-xB)$. We assume that $\log(.)$ is the principal log.; thus we assume that, for every eigenvalue $\lambda$ of $B$ and $x\in I$, $1-x\lambda\notin (-\infty,0]$.

When $x$ is small, $g(x)=-xB-(xB)^2/2-(xB)^3/3-\cdots$ and $g'(x)=-B(I+xB+(xB)^2+(xB)^3+\cdots)=-B(I-xB)^{-1}$. By extending the equalities between holomorphic functions, $g'(x)=-B(I-xB)^{-1}$ is valid for any $x\in I$.

Now $f'(x)=-tr(Ag'(x))=tr(AB(I-xB)^{-1})$ and we are done.

EDIT. Answer to mark. 1. $g'(x)$ and $-B(I-xB)^{-1}$ are analytic functions; there are equal in a neighborhood of $0$; then they are equal on whole domain of definition -if it is connected-.

  1. The set $E=\{1-x\lambda; x\in I,\lambda\in sp(B)\}$ is included in $n$ straight lines going through $1$. Then, there is a half line $D$ from $0$ that does not intersect $E$. Finally, choose the $\log(.)$ associated to $D$. It remains the conditions $1-x\lambda\not= 0$.
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  • $\begingroup$ Many thanks for this, very useful indeed. I think I need some help to understand completely though. (i) From your second paragraph, I can see $g'(x)=-B(I-xB)^{-1}$ for small $x$. What is the argument needed to show this actually holds for any $x\in I$? (ii) What can be done when $x\in I$ and $1-x\lambda\in (-\infty,0]$? Can a solution for the integral still be given? $\endgroup$ – mark Jan 18 '16 at 13:46
  • $\begingroup$ thanks so much for the edit, I feel privileged that you’re explaining this to me. Concerning your answer 1, I can now see that $g'(x)=-B(I-xB)^{-1}$ over any interval containing an interval where the series for $g(x)$ and $g'(x)$ converge. But now suppose the domain $I$ is $(0,b)\cup(b,c)$, where $b\neq0$ is the reciprocal of a real eigenvalue of $B$ ($I-xB$ is singular at $x=b$), and $c$ is some real scalar greater than $b$. Can I show $g'(x)=-B(I-xB)^{-1}$ over a disconnected set $I$ of this type? This must be true, mustn't it? $\endgroup$ – mark Jan 19 '16 at 10:41
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    $\begingroup$ We consider the $\log(.)$ associated to the half line $D$ (that is not parallel to $Ox$). Then the condition associated to the eigenvalue $\lambda=1/b>0$ is $x\notin \Delta$ a half line from $b$ that is parallel to $D$ (here $x$ is seen as a complex number). $g'(x)=-B(I-xB)^{-1}$ is valid for small complex numbers $x$. Since $\mathbb{C}\setminus \Delta$ is connected, the equality is valid in whole the previous set and in particular on $(0,b)\cup (b,c)$. $\endgroup$ – loup blanc Jan 19 '16 at 16:33
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    $\begingroup$ @ mark , the indefinite integral is not real (there is the constant $i\pi$); yet, the definite integral $\int_a^b(.)dx$ is real. $\endgroup$ – loup blanc Jan 23 '16 at 9:52
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    $\begingroup$ @ mark . 1) A solution is $-tr(\log(I-xA))$. 2) $tr(\log(I-xA)=\log(\det(I-xA))$ is false. (Have a look when several eigenvalues of $I-xA$ are $<0$). 3) I think that, if $A$ is real, then a primitive has the form $\log(|\det(I-xA)|)$; yet beware, $x$ must go through a real SEGMENT where the conditions $1-x\lambda\not=0$ are always satisfied. $\endgroup$ – loup blanc Jan 26 '16 at 22:09

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