Constructing a continuous matrix valued function

Question

Given $d<k$. Let ${\cal M}_{d\times k}(\mathbb{R})$ denotes the set of all $d\times k$ real matrices and suppose that $H:\mathbb{R}^k\rightarrow {\cal M}_{d\times k}(\mathbb{R})$ 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 $K:\mathbb{R}^k\rightarrow {\cal M}_{k\times (k-d)}(\mathbb{R})$ 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 $K(x_0)\in {\cal M}_{k\times (k-d)}(\mathbb{R})$ 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.

Answer 1

I got an answer that requires an additional hypothesis:

• $\forall x$, every principal submatrix of $H(x)$ has to be non singular.

Here, we define the principal submatrices of a generic matrix $A\in\mathbb{R}^{d \times k}$ as the matrices $A^{(1)},\ldots,A^{(d)}$ given by $A^{(m)}\in\mathbb{R}^{m \times m}$ and $[A^{(m)}]_{i,j} = [A]_{i,j}$ for each $m = 1,\ldots,d$.

Now for each $x$ you compute the row echelon form of $H(x)$ by means of gaussian elimination, resulting into the matrix $E(x)$. This process is continuous because you don't need pivoting, thanks to the initial hypothesis. (http://en.wikipedia.org/wiki/Row_echelon_form)

Now, you have $$E(x)= \begin{bmatrix} 1 & * & * & * & * & \cdots & * \\ & 1 & * & * & * & \cdots & * \\ & & \ddots & \ddots & & & \vdots\\ & & & 1 & * & \cdots & * \end{bmatrix}$$
where "$*$" are generic non zero elements. Consider now the linear system $E(x)y = 0$, of the form $$ \begin{cases} y_1 + (\text{combination of $y_2,\ldots,y_k$}) = 0\\ y_2 + (\text{combination of $y_3,\ldots,y_k$}) = 0\\ \vdots \\ y_d + (\text{combination of $y_{d+1},\ldots,y_k$}) = 0 \end{cases} $$ and build your basis $\mathcal{B} = \mathcal{B}(x)$ of the space $\{E(x)y = 0\}$ (that coincides with $\{H(x)y = 0\}$ ) as follows: $$\mathcal{B} = \{b^{(1)},\ldots,b^{(k-d)}\}\subseteq\mathbb{R}^k$$ where, for each $m$, the last $(k-d)$ components of $b^{(m)}$ are given by $$[b^{(m)}]_{d + i} = \delta_{i,m}, \quad \text{for } i = 1,\ldots,k-d$$ and you can compute the first $d$ components of each $b^{(m)}$ by direct substitution on the linear system $E(x)y = 0$.

Again, the whole process is continuous and now you just have to consider the matrix $K(x)$ that has the vectors of $\mathcal{B}$ as columns.