Measurable projection on the null space of a random matrix

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 $$U$$ ?

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.

• 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. Nov 15, 2021 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 $\omega \mapsto (\Sigma(\omega),f(\omega))$. Nov 15, 2021 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. Nov 15, 2021 at 18:30
• @blamethelag Ok but then how do you obtain the map $g$ from this? Its not simply $g\circ f$ no? Nov 15, 2021 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. Nov 15, 2021 at 19:02

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.

• Thank you for your answer. Any ideas on how to proceed from there? Nov 17, 2021 at 12:19
• I made an edit to my post with an attempt. Do you think its ok? Nov 17, 2021 at 14:30