4
$\begingroup$

I am wondering how to integrate the function

$$x \mapsto \operatorname{tr}\bigl\{(\mathbf{A}x+\mathbf{B})^{-1}\mathbf{C}\bigr\}$$

In my case, the matrices $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ are (strictly) positive definite. If $\mathbf{C} = \mathbf{A}$ it is easy to see that

$$\ln\det(\mathbf{A}x+\mathbf{B})$$

is an antiderivative. But what if $\mathbf{C} \neq \mathbf{A}$?

Thanks,
jens

  • 2
    This might help: $$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx=\mathrm{tr}\left( \int (Ax+B)^{-1}dx\, C\right). $$2012-07-01
  • 1
    @anon : Perhaps you should turn this in an answer since it pretty much says everything!2012-07-01
  • 0
    Thanks a lot anon! I'm just not used to handling matrix logarithms. Is that the reverse of a matrix exponential $\mathrm{exp}(\mathbf{A})$? What still intrigues me is that, while at first glance your derivation makes sense to me, the antiderivative turns out to be a trace of three matrices: $\log(\cdot)$, $\mathbf{A}^{-1}$ and $\mathbf{C}$. The latter two are positive definite, while the $\log(\cdot)$ is not even Hermitian in general, as far as I understood. I'd be tempted to say that if you choose the values of $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$, $x$ at leisure, that trace of three matr2012-07-01
  • 0
    The account used to post the question here was unregistered, and it appears you were unable to get back into the account. Eric Naslund did some mod voodoo. Anyway: log is the inverse of exp and has a power series expansion equivalent to the real case (with convergence technicalities, naturally). There are also very interesting and deep issues dealing with noncommutativity relevant to Lie theory (see the BCH formula). If you plug in all real values you should get a real trace, otherwise I'm not sure what I can say.2012-07-01

1 Answers 1

1

By the linearity of integration, trace, and matrix multiplication, we can pull $\int$ inside:

$$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx=\mathrm{tr}\left( \int (Ax+B)^{-1}dx\, C\right).$$

I originally thought $\log(Ax+B)\;A^{-1}$ would be the antiderivative for this, but oenamen helpfully points out this can't be said in full generality ($\log XY=\log X+\log Y$ does not necessarily hold if the matrices $X$ and $Y$ don't commute!); instead, we need a special $B^{-1}$ factor. We have:

$$\frac{d}{dx}\log(I+xU) =\quad U(I+xU)^{-1} =(I+xU)^{-1}U.$$

Hence, applying the above alongside $X^{-1}Y^{-1}=(YX)^{-1}$:

$$\begin{array}{c l} \frac{d}{dx}\log \big(B^{-1}(Ax+B)\big) & =\frac{d}{dx}\log(I+xB^{-1}A) \\ & = (I+xB^{-1}A)^{-1}B^{-1}A \\ & = \big(B\cdot(I+B^{-1}Ax)\big)^{-1}A \\ & = (Ax+B)^{-1}A. \end{array}$$

Therefore we conclude

$$\int \mathrm{tr}\big( (Ax+B)^{-1}C\big)dx =\mathrm{tr}\left(\log\big(I+xB^{-1}A)\; A^{-1}C\right)+const. $$

  • 0
    Very nice. What happened to the comments?2012-07-01
  • 0
    @oenamen This is a second answer; the first version I deleted. :<2012-07-01
  • 0
    Ah, for a second there I thought you had some superuser privileges!2012-07-01