It's well known that when the elements of an $n \times n$ matrix $A$ are chosen independently from e.g. $U[0,1]$ distributions, then with probability $1$ the matrix $A$ will be injective (indeed, invertible). Given the Hilbert space $L^2(\mathbb{R}^d)$, a Hilbert-Schmidt kernel $k \in L^2(\mathbb{R}^d \times \mathbb{R}^d)$ and the corresponding Hilbert Schmidt operator $T$ has a natural representation as an infinite matrix,

$k(x,y) = \sum_{i,j} \lambda_{ij} e_i(x) e_j(y)$

$(Tf)(x) = \int f(y)k(x,y) \, \mathrm{d}y$

where $\sum_{ij} \lambda_{ij}^2 < \infty$ and $\{e_i(x)\}$ is a (fixed) orthonormal basis of $L^2(\mathbb{R}^d)$. Suppose that the $\lambda_{ij}$ are drawn independently from continuous distributions supported on $[0,\overline{\lambda}_{ij}]$ where $\sum_{ij} \overline{\lambda}_{ij}^2 < \infty$. It seems to me that $T$ should generically (i.e. with probability $1$) be injective, noting that any inverse would not be bounded. This corresponds with any row of the infinite matrix $\{\lambda_{ij}\}$ not being in the linear span of the other rows. I'm curious as to whether there has been any work in this direction?

There was a similar question asked here, but that was more on constructing a specific ``random operator" along the lines above than proving that invertibility is common among them: Is there a good notion of "random bounded linear map" on a separable Hilbert space?

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    $\begingroup$ Intuitively, it seems clear that conditional on rows $1,2,\dots,n-1$, the $n$th row has zero probability to lie in the $n-1$-dimensional subspace spanned by the previous rows, since that's a finite-dimensional subspace in an infinite-dimensional space. By induction and countable additivity we should get that the matrix is a.s. injective. $\endgroup$ Feb 17 '18 at 20:41
  • $\begingroup$ Hi Nate, it certainly seems that a row lies in the linear span of finitely many other rows of the matrix with $0$ probability; however, injectivity of the operator would also require that any row does not equal the linear combination of infinitely many of the other rows, which to me is less clear. $\endgroup$
    – Ikebf
    Feb 17 '18 at 20:44
  • $\begingroup$ Oh yeah, good point. $\endgroup$ Feb 17 '18 at 20:45
  • $\begingroup$ Just to clarify: are you fixing the choice of basis $e_i$, and then choosing the $\lambda_{ij}$ randomly? $\endgroup$
    – Yemon Choi
    Feb 17 '18 at 21:09
  • $\begingroup$ @YemonChoi, yes that's correct. $\endgroup$
    – Ikebf
    Feb 17 '18 at 21:11

It depends on the distribution. For instance, if we restrict our attention to the case where $\lambda_{ij}$ is uniform on $[0, \bar{\lambda}_{ij}]$, then we can choose $\bar{\lambda}_{ij}$ so as to make $T$ a.s. injective, but we can also chose them so that this is not the case.

By taking transposes, it's equivalent to ask whether $T$ has dense image.

For one direction, let's begin by seeing how to choose $\bar{\lambda}_{ij}$ so that $e_1$ is almost surely in the closure of the image. Let $a_j$ be your favorite square-summable sequence of positive numbers, and set $\bar{\lambda}_{1j} = a_j$. Then select $\bar{\lambda}_{ij}$, $i \ne 1$, so that for each $j$ we have $\sum_{i=2}^\infty |\bar{\lambda}_{ij}|^2 < \frac{1}{j} a_j^2$. Generate a matrix $T$ with the resulting distribution. Almost surely, we have $\lambda_{1j} \le \frac{1}{2} a_j$ for infinitely many values of $j$. For such a $j$, we have $T(\lambda_{1j}^{-1} e_j) = e_1 + \lambda_{1j}^{-1} \sum_{i=1}^\infty \lambda_{ij} e_i$, so $$\begin{align*}\|T(\lambda_{1j}^{-1} e_j) - e_1\|^2 &= \lambda_{1j}^{-2} \sum_{i=2}^\infty |\bar{\lambda}_{ij}|^2\\ &\le \frac{1}{j} \lambda_{1j}^{-2} a_j^2 \\ &\le \frac{4}{j}\end{align*}$$ Thus $e_1$ is in the closure of the image.

Now you can construct similar matrices for which $e_2, e_3, \dots$ are in the closure of the image, and shuffle together their columns so that the image of the resulting matrix contains the union of their images, and thus is dense.

In the other direction, choose $a_j$ as before and set $\bar{\lambda}_{j+1, j} = a_j$. This matrix is going to look somewhat like a right shift. Choose another sequence $c_j$ so small that $\prod_{j=1}^\infty (1 - 3c_j) > 0$, and choose $\bar{\lambda}_{ij}$, $i \ne j+1$, so that $\sum_{j=1}^\infty (c_j a_j)^{-1} \sum_{i \ne j+1} |\bar{\lambda}_{ij}|^2 \le 1$. Now by splitting up the matrix $T$ into the entries on and off the shifted diagonal, we can write $$T = SUA + MCA$$ where $$\begin{align*}A &= \operatorname{diag}(a_1, a_2, \dots) \\ C &= \operatorname{diag}(c_1, c_2, \dots) \\S &= \text{right shift} \\ U &= \operatorname{diag}(U_1, U_2, \dots), && U_j \text{ iid } \mathrm{unif}(0,1) \\ \|M\| &\le \|M\|_{HS}\le 1. \end{align*}$$ The $c_j$ were chosen so that with positive probability, we have $U_j \ge 3c_j$ for all $j$. On this event, for any vector $x$ we have $\|Ux\|^2 = \sum_j |U_j x_j|^2 \ge \sum_j |3 c_j x_j|^2 = (3 \|Cx\|)^2$. Since $S$ is an isometry, we have $\|SUAx\| = \|UAx\| \ge 3 \|CAx\|$. And $\|MCAx\| \le \|CAx\|$ since $\|M\| \le 1$. Hence $\|Tx\| \ge \|SUAx\| - \|MCAx\| \ge 2 \|CAx\|$ for any $x$.

On the other hand, we have $$|\langle Tx, e_1 \rangle| = |\langle MCAx, e_1\rangle| \le \|MCAx\| \le \|CAx\|.$$ Combining these gives $|\langle Tx, e_1 \rangle| \le \frac{1}{2} \|Tx\|$ for all $x$. In particular, whenever $\|Tx \| = 1$ we have $\|Tx - e_1\|^2 = 2 - 2 \langle Tx, e_1 \rangle \ge 1$. So $e_1$ is not in the closure of the range of $T$.

  • $\begingroup$ Thanks for the great answer; it's interesting that the answer to the question depends on the choice of $\overline{\lambda}_{ij}$, but intuitive in that with appropriate choices we can make the operator look enough like identity or right shift (probability-wise). $\endgroup$
    – Ikebf
    Feb 19 '18 at 1:09

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