Dgdrxrt

We have as a known data matrices $A$,$T$.

We want to find $S$ that $A=ST$.
What I would do is multiply $T^-1$ from right side.

$AT^-1=S$

And here we have $S$, but what if $det(T)=0$ so matrix $T^-1$ does not exists.
Does it implify that searched $S$ also does not exists?

The matrix $S$ may exist, but it is also possible that it doesn't exist. If, for instance, $det Aneq0$, then, since $det T=0$, you can be sure that it doesn't exist.

Suppose $S$ exists and we're dealing with square matrices. Then if $Tv=0$, we also have $Av=0$.

Since the null space of $T$ is contained in the null space of $A$, the rank-nullity theorem tells us that the rank of $T$ is greater than or equal than the rank of $A$.

Thus $S$ cannot exist if the rank of $A$ is greater than the rank of $T$.

Actually, the matrix $S$ exists if and only if the null space of $T$ is contained in the null space of $A$. It's perhaps simpler to show it with linear maps.

We have linear maps $fcolon Vto V$ and $gcolon Vto V$ (with $V$ a finite dimensional space) such that $ker gsubseteqker f$. We want to find $hcolon Vto V$ such that $f=hcirc g$.

Take a basis $v_1,dots,v_k$ of $ker g$ and extend it to a basis of $V$. Then it's easy to show that $w_k+1=g(v_k+1),dots,w_n=g(v_n)$ is linearly independent, so we can extend it to a basis $w_1,dots,w_k,w_k+1,dots,w_n$ of $V$.

Now define $h$ on this basis by
$$
h(w_i)=begincases
0 & 1le ile k \[4px]
f(v_i) & k+1le ile n
endcases
$$
Since, for every $i$, we have $h(g(v_i))=f(v_i)$, we can conclude that $hcirc g=f$.

Now take for $f$ and $g$ the linear maps induced by $A$ and $T$ respectively: $f(v)=Av$ and $g(v)=Tv$; for $S$ use the matrix of $h$ with respect to the standard basis.

Of course the condition that the null space of $T$ is contained in the null space of $A$ is satisfied if $T$ is invertible.