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Let $H$ be a separable Hilbert space with inner product $\langle,\rangle$, let $\{e_k\}_{k=1}^\infty$ be an orthonormal basis of $H$, and let $A: H\to H$ be a symmetric, positive definite and invertible operator. Let $u\in H$ and $n\in \mathbb{N}$, I aim to evaluate the following integral $$ \int_{\mathbb{R}^n}\exp\left(-\frac{1}{2}\left\langle A \left(\sum_{k=1}^n x_k e_k-A^{-1}u\right), \sum_{k=1}^n x_k e_k -A^{-1}u\right\rangle \right)\, dx_1\ldots dx_n. $$ It would be ideal if it can be computed for general $u$, but I am happy to assume $u$ lies in the subspace $H_n$ spanned by $\{e_i\}_{i=1}^n$.

Do I have to assume $A$ or $A^{-1}$ preserves $H_n$?

My attempt:

To evaluate the integral, I try to rewrite the integral as an integral of Gaussian density. Note that \begin{align*} &\left\langle A \left(\sum_{k=1}^n x_k e_k-A^{-1}u\right), \sum_{k=1}^n x_k e_k -A^{-1}u\right\rangle \\ &=\left\langle \sum_{k=1}^n x_k A e_k- u, \sum_{k=1}^n x_k e_k -A^{-1}u\right\rangle \\ &=x^\top \mathbf{A} x- 2x^\top \mathbf{b} +\langle u, A^{-1} u\rangle, \end{align*} where $\mathbf{A}=(A_{ij})_{i,j}\in \mathbb{R}^{n\times n}$ and $\mathbf{b}=(b_i)_{i=1}^n\in \mathbb{R}^{n}$ are defined as $$ A_{ij} = \langle Ae_i, e_j\rangle, \quad b_i = \langle u, e_i\rangle. $$ As $\mathbf{A}$ is a positive definite matrix, it is invertible. Then \begin{align*} x^\top \mathbf{A} x- 2x^\top \mathbf{b} +\langle u, A^{-1} u\rangle = (x-\mathbf{A}^{-1} \mathbf{b})^\top \mathbf{A} (x-\mathbf{A}^{-1} \mathbf{b}) - \mathbf{b}^\top \mathbf{A}^{-1} \mathbf{b}+ \langle u, A^{-1} u\rangle. \end{align*} If I could show \begin{align} \tag{1} &\mathbf{b}^\top \mathbf{A}^{-1} \mathbf{b}- \langle u, A^{-1} u\rangle =0, \end{align} then I can rewrite the integral as an integration with a multivariate Gaussian density, which can be computed involving the determinant of $\mathbf{A}^{-1}$. However, I am not sure whether (1) holds or not. Even assuming $u=\sum_{k=1}^n \langle u,e_k\rangle e_k$, we still need to show $$ \sum_{k,l=1}^m \langle u, e_k\rangle \left(( \mathbf{A}^{-1})_{kl} -\langle A^{-1}e_k ,e_l\rangle \right) \langle u, e_l\rangle=0. $$ It is unclear to me whether the identity holds, since typically $A$ may not map $H_n$ into $H_n$.

This seems to be related to the matrix representation of an inverse operator, which has been asked here. However, it assumes $A$ maps $H_n$ into itself, and hence we can use the same basis for the range of $A$.

John
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1 Answers1

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Your integral can be written, using that $A$ is positive, as $$ I=\int_{\mathbb R^n}\exp(-\frac12\,\|A^{1/2}(x-A^{-1}u)\|^2)\,dx $$ If you assume that $A$ maps $H_n$ into itself, the change of variable $$v=A^{1/2}(x-A^{-1}u)=A^{1/2}x-A^{-1/2}u$$ gives you the factor $|\det A^{-1/2}|=1/\sqrt{\det A}$, and so $$ I=\frac1{\sqrt{\det A}}\,\int_{\mathbb R^n}\exp(-\frac12\,\|x\|^2)\,dx=\frac{(2\pi)^{n/2}}{\sqrt{\det A}}. $$ In the general case we may write $$\langle Ax,x\rangle=\langle APx,Px\rangle=\|A^{1/2}Px\|^2=\|(PAP)^{1/2}x\|^2,$$ where $P$ is the orthogonal projection onto $H_n$. Since $A$ is invertible, the operator $A^{1/2}P$ has rank $n$, and so does $PAP$, which is then invertible as an operator on $H_n$; and a fortiori $(PAP)^{1/2}$ is invertible, too. Thus $$ I=\frac{(2\pi)^{n/2}}{\sqrt{\det (PAP)}}. $$ There is no obvious formula for $\det PAP$ in terms of $A$ (and note that $\det A$ has no meaning for arbitrary $A$).

Edit: a bit more detail. In general one has $$\ker T^*T=\ker T$$ (because $T^*Tx=0$ implies $\|Tx\|^2=\langle T^*Tx,x\rangle=0$). In the finite-dimensional case this implies that $T^*T$ has the same rank as $T$. So $PAP=(A^{1/2}P)^*A^{1/2}P$ has the same rank as $A^{1/2}P$. Now $PAP:H_n\to H_n$ has full rank and is henceforth invertible.

Martin Argerami
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  • For the general case, I understand $ PAP$ is of rank at most $n$, as it maps into $H_n$. But may you elaborate more why $PAP:H_n \to H_n$ is invertible? – John Mar 19 '23 at 10:18
  • Also I guess the identity $\langle Ax,x\rangle =\langle APx Px\rangle$ only holds with $x\in H_n$? But since $ A^{-1} u$ may not be in $H_n$, can we still apply the formula to rewrite the integral? – John Mar 19 '23 at 10:24
  • I've edited a clarification in. At first sight, you are right that where $A^{-1}u$ lives might cause problems. But you said you could assume $u\in H_n$. – Martin Argerami Mar 19 '23 at 10:29
  • Thank you for clarifying the first point. I am happy to assume $u\in H_n$. But why does $u\in H_n$ help for the second issue? It is still unclear that $A^{-1}u\in H_n$, right? – John Mar 19 '23 at 10:32
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    Well, the obvious thing is that the issue disappears if $A^{-1}u\in H_n$. Other than that, I don't see an obvious path. I'll have to think about it. – Martin Argerami Mar 19 '23 at 10:41
  • Thanks. If I assume $A$ maps $H_n$ to $H_n$, does it imply $A^{-1}$ will map $H_n$ to $H_n$? It seems that you have used this fact implicitly in the first case to derive $A^{-1}u\in H_n$ under the assumption that $u\in H_n$. – John Mar 19 '23 at 15:54
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    If $A$ maps $H_n$ into $H_n$, because $\dim H_n<\infty$ and $A$ is injective, the restriction of $A$ to $H_n$ is invertible. If now $x\in H_n$, then $x=Ay$ for some $y\in H_n$. Thus $A^{-1}x=A^{-1}(Ay)=y\in H_n$. – Martin Argerami Mar 19 '23 at 20:32