1

Consider a random $n\times n$ real matrix $\Sigma$ on a measure space $(\Omega,\mathcal F)$. Let $\ker(\Sigma)$ denote its random null space. If $f:(\Omega,\mathcal F)\to\mathbb R^n$ is a Borel measurable map, can one show the existence of a Borel measurable map $\mathcal g:(\Omega,\mathcal F)\to\mathbb R^n$ such that, for all $\omega\in\Omega$,

$$g(\omega)= P_{\ker(\Sigma(\omega))}\,f(\omega)$$

where $P_{U}\,v$ denotes the orthogonal projection of $v$ onto the subspace $U$ ?

Thanks for your help.

EDIT: I made an attempt below. Any feedback is very appreciated.

Let $M_n$ denote the space of $n\times n$ real matrices endowed with the Frobenius matrix norm. Let $\mathcal B(M_n)$ denote the Borel $\sigma$-algebra on $M_n$. Since $M_n$ is homeomorphic with $\mathbb R^{n^2}$ there is a bijection between $\mathcal B(M_n)$ and $\mathcal B(\mathbb R^{n^2})$.

I will try to show that the map $(\Sigma,v) \mapsto P_{\ker(\Sigma)}v$ from $(M_n\times \mathbb R^n,\mathcal B(M_n) \otimes \mathcal B(\mathbb R^{n^2}))$ to $(\mathbb R^n,\mathcal B(\mathbb R^n))$ is measurable. The result will then follow by composing with the measurable map $\omega\mapsto (\Sigma(\omega),f(\omega)$).

The idea is to apply a principle of measurable choice as stated here. To this end let $X=M_n\times \mathbb R^n $, $Y=(\mathbb R^n)^3$ and define

$$E=\bigg\{\big(\Sigma,v,u,w,x\big)\in X\times Y: v=u+w,u=\Sigma^\top x, \Sigma w=0 \bigg \}$$

Since a finite product of separable spaces is separable, we have that $X,Y$ are separable metric spaces. Using the sequential characterization of a closed set in a metric space and continuity of matrix multiplication, we see that $E$ is a closed subset of $X\times Y$. Since $X\times Y$ is homeomorphic with $\mathbb R^{n^2+4n}$, it is $\sigma$-compact. Closed subsets of $\sigma$-compact spaces are also $\sigma$-compact, so $E$ is $\sigma$-compact.

The principle of measurable choice therefore provides us with a Borel function $\varphi :\pi_X(E)\to Y$ whose graph is contained in $E$, where $\pi_X$ denotes the projection on $X$.

Note that $\pi_X(E)=X$, because $\mathbb R^n = \ker(\Sigma) \oplus \text{range}(\Sigma^\top)$. Moreover we have

$$\varphi_1(\Sigma,v)=P_{\text{range}(\Sigma^\top)}v$$

$$\varphi_2(\Sigma,v)=P_{\ker(\Sigma)}v$$

where $\varphi_1,\varphi_2$ denote the first and second component functions of $\varphi$, which are Borel measurable since $\varphi$ is.

The final thing to check is that $\mathcal B(X)=\mathcal B(M_n) \otimes \mathcal B(\mathbb R^{n^2})$, but this follows from the fact that $M_n$ and $\mathbb R^{n^2}$ are separable.

Alphie
  • 4,740
  • If I understood correctly your question, you are wondering if $\omega \mapsto P_{ker(\Sigma(\omega))} f(\omega)$ is measurable for any $f$ measurable. I think this is equivalent to ask $\omega \mapsto P_{ker(\Sigma(\omega))}$ measurable. – blamethelag Nov 15 '21 at 18:12
  • @blamethelag Yes, but how is your map $\omega \mapsto P_{ker(\Sigma(\omega))}$ defined? I think it is sufficient to show that the map $(\Sigma,v) \mapsto P_{ker(\Sigma)}v$ is measurable, for then we can obtain the desired map by compounding with the measurable map $\omega \mapsto (\Sigma(\omega),f(\omega))$. – Alphie Nov 15 '21 at 18:24
  • The map is defined as matrix or linear map valued endowing the codomain of the Borel sigma algebra corresponding. Fixing a basis and seeing $P_{ker(\Sigma(\omega))}$ as a matrix seems the easiest option. – blamethelag Nov 15 '21 at 18:30
  • @blamethelag Ok but then how do you obtain the map $g$ from this? Its not simply $g\circ f$ no? – Alphie Nov 15 '21 at 18:33
  • $g \circ f$ makes no sens. In what you wrote $g$ is necessarily $\omega \longmapsto P_{ker(\Sigma(\omega))}f(\omega)$ so the notation $g$ is useless. – blamethelag Nov 15 '21 at 19:02
  • @blamethelag Yes sorry I meant the composition of $f$ with your function. Anyway you have an idea how to proceed? – Alphie Nov 15 '21 at 20:23
  • Maybe note that the kernel of $\Sigma$ is the orthogonal complement of the range of $\Sigma^T$. Then you can apply Gram-Schmidt (with treating linear dependencies reasonably) to construct the projection on the range of $\Sigma^T$ in a measurable way. Use this to show that the projection on the kernel is also measurable. – PhoemueX Nov 16 '21 at 20:44
  • @PhoemueX I made an edit to my post with an attempt. Do you think its ok? – Alphie Nov 17 '21 at 14:25

1 Answers1

1

Let $(1)$ be the assertion :

For all $f : \Omega \longrightarrow \mathbb R^n$ measurable, there exists $g : \Omega \longrightarrow \mathbb R^n$ measurable such that $$ \forall \omega \in \Omega,~~~P_{ker(\Sigma(\omega))}f(\omega) = g(\omega). $$

and $(2)$

The application $\omega \longmapsto P_{ker(\Sigma(\omega))}$ is measurable.

Let me show $(1)$ and $(2)$ are equivalent.

$\Longrightarrow$ : Assume $(1)$ is true, denote $(e_k)$ the canonical basis of $\mathbb R^n$, then for all $k = 1,...,n$, applying $(1)$ to $f(\omega) = e_k$ yields that $\omega \longmapsto P_{ker(\Sigma(\omega))}e_k$ is measurable. Then by definition of the product sigma algebra the application $$ \omega \longmapsto (P_{ker(\Sigma(\omega))}e_k)_{1 \leq k \leq n} $$ is measurable from $\Omega$ to $(\mathbb R^n)^n$. The application that to a family of vectors assigns the corresponding matrix is measurable (it is continuous) so we have $(2)$.

$\Longleftarrow$ : Assume $(2)$, then take $f$ measurable, the matrix-vector product being measurable (it is continuous) you get that $$ \omega \longmapsto P_{ker(\Sigma(\omega))}f(\omega) = ((M,v) \longmapsto Mv) \circ (\omega \longmapsto (P_{ker(\Sigma(\omega))}, f(\omega))) $$ is measurable.

Then re consider your question, you noticed that $$ \omega \longmapsto P_{ker(\Sigma(\omega))}= (M \longmapsto P_{Ker(M)}) \circ \Sigma $$ so I think the only reasonable way to solve your problem is either to show that $M \longmapsto P_{Ker(M)}$ is measurable or either to provide an example of $\Sigma$ which shows $\omega \longmapsto P_{ker(\Sigma(\omega))}$ is not measurable.

blamethelag
  • 2,007