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

  • $\begingroup$ 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

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 $1-x\lambda\not= 0$.
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @ 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
  • 1
    $\begingroup$ @ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.