1
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ @JIamprong The Grahm Schmidt process is a continuous process. The following link may help you indirectly math.stackexchange.com/questions/288912/… $\endgroup$ Jan 21, 2014 at 16:02
  • 1
    $\begingroup$ Before using Gram–Schmidt, I would notice that the pointwise kernel of the map $H$ defines a vector bundle over the source of $H$. Since the source of $H$ is (paracompact Hausdorff and) contractible, then that vector bundle is trivializable. Now apply the Gram–Schmidt orthogonalization process. $\endgroup$ Jan 21, 2014 at 16:12
  • $\begingroup$ @RicardoAndrade: I don't know about the source of $H$. Could you explain me more detail. I am not really familiar with this. Thanks $\endgroup$
    – Jlamprong
    Jan 21, 2014 at 16:16
  • $\begingroup$ The source of $H$ means its domain, i.e. $\mathbb{R}^k$ in your case. $\endgroup$ Jan 21, 2014 at 16:17
  • $\begingroup$ SO, what does "the vector bundle is trivializable" means? Btw, can we construct such matrix? $\endgroup$
    – Jlamprong
    Jan 21, 2014 at 16:20

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.