# Can we show a matrix always exists to satisfy these properties

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?