Vrftsjtryk

Given $d<k$. Let ${cal M}_{dtimes k}(mathbb{R})$ denotes the set of all $dtimes k$ real matrices and suppose that $H:mathbb{R}^krightarrow {cal M}_{dtimes k}(mathbb{R})$ is a continuous matrices-valued function such that $H(x)$ is full rank for every $x in mathbb{R}^k$.

I'd like to construct a continuous function $K:mathbb{R}^krightarrow {cal M}_{ktimes (k-d)}(mathbb{R})$ such that $K(x)$ is full rank and
begin{equation}
H(x)K(x)=0, quad forall x in mathbb{R}^k.
end{equation}
Can we do that?

I've tried defining $K$ as follows: for every $x_0$ define $K(x_0)$ by such matrix with columns are all element in the basis of the subspace ${y in mathbb{R}^k :H(x_0)y=0}$. Of course $K(x_0)in {cal M}_{ktimes (k-d)}(mathbb{R})$ since ${y in mathbb{R}^k :H(x_0)y=0}$ has dimension $k-d$. But the problem was on the continuity because we can choose arbritary basis of the above subspace. Can anyone give advice in constructing $K$? Thanks in advance.

Here are some thoughts on this question that are too lengthy for a comment. First, if you know any procedure that generates orthogonal vector for a given system of vectors and this procedure depends continuously on vectors of set (and depends only on vectors of this set) — it'll work here. You take vectors which are rows of $H(x)$, generate orthogonal vector $K_1$ and obtain matrix $( H; vert ; K_1)^{T}$ which is continuous w.r.t. to $x$. Then you repeat procedure and obtain vector $K_2$, append it again and repeat until you construct matrice $K(x) = (K_1, K_2, dots, K_{k-d})$. I've tried to find such procedure, but hasn't succeed yet.

Also there's a further development of idea from comments, but it has a flaw. I'll use following notation:

• each point $x in mathbb{R}^k$ is equipped with $d$ vectors that form columns of matrix $hat{H}(x) = H^{T}(x)$, $H(x) in mathbb{R}^{d times k} $, $hat{H}(x) in mathbb{R}^{k times d} $;




• by $E$ denote the identity matrix, $E in mathbb{R}^{ktimes k}$;



• by $Gamma (x)$ denote Gramm matrix, $Gamma(x) = H(x)cdot H^{T}(x)$, $Gamma(x) in {mathbb R}^{d times d}$.

Then you do the following. For each point compute matrix $hat{K}(x) = E - hat{H}(x) cdot Gamma^{-1}(x) hat{H}^{T}(x)$. This transformation just takes a standard basis at each point $x$ (it corresponds to matrix $E$), subtracts from each basis vector its orthogonal projection on $hat{H}(x)$ and makes a set of vectors from $hat{H}^{perp}(x)$. Matrix $hat{K}(x)$ depends continuously on $x$ (multiplications/additions are continuous, inversion is continuous too). Also, it satisfies $hat{H}(x)hat{K}(x) equiv 0$. The only problem (and this is the flaw that I've mentioned) is that $hat{K}(x)$ is from $mathbb{R}^{k times k}$ instead of $mathbb{R}^{k times (k-d)}$; but it surely has rank $(k-d)$. The problem that I'm struggling here is that I still don't know a way how to continuously "extract" some basis from set of columns of matrix $hat{K}(x)$.

EDIT: Probably, that's not always possible. I wonder if OP's original can be solved by any other method.