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Is the following integral a convergent integral? Can we compute it, precisely?

$$\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu $$

Here $\mu$ is the usual measure of $M_{n}(\mathbb{R})\simeq \mathbb{R}^{n^{2}}$?

So $\mu$ can be counted as $\mu=\prod_{i,j} da_{ij}$

Note: If this integral would be convergent , either in Lebesgue or in Riemann sense, then it would be equal to a scalar matrix. Because for every invertible matrix $P$ we have:

$P^{-1}(\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu) P= \int_{M_{n}(\mathbb{R})} e^{-(P^{-1}AP)^{2}}d\mu=\int_{M_{n}(\mathbb{R})} e^{-A^{2}}d\mu$ since the mapping $A\mapsto P^{-1}AP$ is a measure preserving and volum preserving linear map.Now we apply the change of coordinate formula for integral.

Ali Taghavi
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  • What is $A^2$ : a matrix or a number ? – Thomas Dec 15 '15 at 16:41
  • @Thomas It is a matrix, as the integral is taken over $"M_{n}(\mathbb{R})"$. – Ali Taghavi Dec 15 '15 at 16:45
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    Remember that for matrices, $A^2$ can have negative eigenvalues, think about a neighboorhod of a rotation matrix, or Pauli spin matrices. So your integral vill not converge to 0 on rays, the same way as $e^{-x^2}$, and the whole thing will not be absolutely convergent. – mlu Dec 15 '15 at 17:10
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    @mlu Do the matrices with real entries and eigenvalues $a+bi$ with $a^2-b^2<0$ have positive measure? I could believe it but I don't see an obvious proof. The point about rays is not a proof because they have measure zero. – Ian Dec 15 '15 at 17:27
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    Without having time to check the exact details right now, a small enough ball around $\left(\begin{matrix} 0 & -1\1 & 0\end{matrix}\right)$ should all have squares having eigenvalues with negative real part (by contiouity). then scale by a factor r to create a cone with positive measure. It is a nice little problem. – mlu Dec 15 '15 at 18:52
  • I have turned mlu's comment into a detailed answer. It is indeed a bit pretty in 2D. – Ian Dec 15 '15 at 19:53

1 Answers1

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With $n=2$, we can look at matrices $A$ such that $A^2$ has eigenvalues with negative real part. When $T(A)^2-4\Delta(A)<0$, the real part of the eigenvalues of $A^2$ is $\frac{T^2}{4}-\frac{4\Delta-T^2}{4}=\frac{2T^2-4\Delta}{4}$. So $A^2$ has eigenvalues with negative real part provided $T^2<2\Delta$ (a stronger condition than $T^2<4\Delta$). One of the simplest matrices with this property is

$$Q=\begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}.$$

Now consider a small perturbation of $Q$:

$$\begin{bmatrix} \delta_1 & -1+\delta_2 \\ 1+\delta_3 & \delta_4 \end{bmatrix}$$

The inequality now reads

$$\delta_1^2 + 2 \delta_1 \delta_4 + \delta_4^2 < 2 \delta_1 \delta_4 - 2(-1+\delta_2)(1+\delta_3).$$

Equivalently:

$$\delta_1^2+\delta_4^2<2+2\delta_3-2\delta_2-2\delta_2 \delta_3.$$

In view of this inequality we can consider the hypercube $H$ with "radius" $1/12$ around $Q$. If $A \in H$ then $T^2-2\Delta<-1$ and so $\frac{2T^2-4\Delta}{4}<-1/2$. Thus the eigenvalues of $A^2$ will have real part less than $-1/2$. Also the volume of $H$ is $(1/6)^4>0$.

Now the integral of $e^{-A^2}$ over the set of all scalar multiples of elements of $H$ cannot converge absolutely.

Ian
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  • +1 thanks for your interesting answer. your answer shows that this integral is not convergent in the Lebesgue sense. What about in the Riemann sense? that is the limit of the integral over discs when the ratio goes to infinity? – Ali Taghavi Dec 16 '15 at 21:52