How can I prove the existence of $X$ for the matrix equation $W^{*}XY = M$ for all $M$?

Question

Lemma:

Suppose $W$ and $Y$ are matrices satisfying $ker(W)=0$ and $ker(Y)=0$. Then for every matrix $M$ there exists a solution $X$ to the equation

$$ W^*XY = M. $$

Attempt:

Since $ker(W)=0$, we have a unique solution $x$ to the equation $Wx = b$ and the inverse $W^{-1}$ exists.

Similarly, the inverse $Y^{-1}$ exists. If $W^{-1}$ exists, we know $(W^*)^{-1}$ exists as well since $(WW^{-1})^* = I^*$.

Then, for any matrix $M$ (with appropriate dimension), there exists $X$ and $X = (W^*)^{-1}MY^{-1}$.

Is this proof sound? If not, how should I prove the given statement?

Edit: There is no explicit mentioning of these matrices being square and so I think my approach is wrong. I am not sure how to prove it otherwise.

Source: page 209 from the textbook (link: http://read.pudn.com/downloads143/ebook/625700/A%20Course%20In%20Robust%20Control%20Theory.pdf)

Answer 1

A helper statement (please check if you are already familiar with it, if not, I can fill in the gaps):

Let $M$ be an $m\times n$ matrix over any field. If $\ker(M)=0$, then there exists an $n\times m$ matrix $M'$ such that $M'M=I_n$.

Knowing $\ker(W)=0$ and $\ker Y=0$, this means that we can find such matrices $Y'$ for $Y$ (with $Y'Y=I$) and $W'$ for $W$ (with $W'W=I$). Note that $W^*(W')^*=(W'W)^*=I^*=I$. Now, for any matrix $M$ of an appropriate size, and look at this:

$$W^*((W')^*MY')Y=(W^*(W')^*)MY'Y=IMI=I$$

Thus we can take $X=(W')^*MY'$ as a solution of your equation $W^*XY=M$.