1
$\begingroup$

Suppose there are two matrices A and B. A has dimension $K \times D$ and B has dimension $D \times K$.

The two matrices obey the following relationships $A = (B^T B) ^{-1} B^T$

$B = \Sigma A^T (A \Sigma A^T) ^{-1}$ where $\Sigma$ is a matrix that has dimension $D \times D$

And let us define the product $P= BA$ where $P$ has dimension $D \times D$.

I am interested in other pairs of matrices that also have a product equal to $P$.

A simple example would be to introduce an invertible matrix C with dimension $K \times K$. If I define new matrices $X=CA$ and $Y=BC^{-1}$ then it is clear that the product $YX$ will also be equal to $P$.

The question I want to answer is whether it is always possible to find a matrix $C$ such that $X$ and $Y$ will obey the following properties:

$Y^TY=\mathbb{1}$

$X=Y^T$

My Attempt (probably wrong)

I'm not really sure how to go about establishing this. My attempt so far has been to expand the two desired properties in terms of A, B, C which leads to

$\begin{align} Y^TY &= \mathbb{1} \\ (BC^{-1})^T BC^{-1} &= \mathbb{1} \\ (C^T) ^{-1} B^T BC^{-1} &= \mathbb{1} \\ B^TB &= C^TC \qquad \text{(1)} \end{align} $

and

$\begin{align} X &= Y^T \\ CA &= (BC^{-1})^T \qquad \text{(2)} \end{align} $

From (1) it is tempting to set C=B but this is not allowed because the dimensions don't match. Beyond that I haven't made any progress in using (1) and (2) to help establish I can always find such a C. I have however read here that (1) does mean there exists an orthogonal matrix O such that C=OB. Is this of any use?

I'd appreciate some help please!

$\endgroup$

0

You must log in to answer this question.

Browse other questions tagged .