what happens to rank of matrices when singular matrix multiply by non-singular matrix?? I need to find the prove for any rectangular matrix $A$ and any non-singular matrices $P$, $Q$ of appropriate sizes,
rank($PAQ$)= rank($A$)
I know that when singular matrix multiply by non-singular the result will be singular matrix. However, It is not relevant to the rank of the matrix. If A is singular with rank 2, why rank($PAQ$)= 2 not any numbers.

More general fact below. The required result would clearly follow from this. Also remember that the rank of a matrix is the number of linearly independent column vectors.

Multiplying with non-singular matrices keeps rank intact.

Let's do pre-multiplication. Post-multiplication is similar, just switch columns with rows.
So let $A$ be non singular and let rank$(B) = r$. Say $B = \left[B_{01} \enspace \cdots \enspace B_{0n} \right]$, and WLOG assume that the linearly independent columns of $B$ are $\{B_{01}, \cdots, B_{0r}\}$, the first $r$ columns.
Clearly, $AB = \left[A(B_{01}) \enspace \cdots \enspace A(B_{0n}) \right] $. Consider the first $r$ columns $\{A(B_{01}), \cdots, A(B_{0r})\}$ of $AB$, and take their linear combination.
Let $\displaystyle\sum_{i=1}^r \lambda_i \cdot A(B_{0i})=\underline 0 \implies A\left( \displaystyle \sum_{i=1}^r \lambda_i \cdot B_{0i}\right)=\underline 0 \implies \sum_{i=1}^r \lambda_i \cdot B_{0i}=\underline 0\enspace,$ as $A$ is non singular and the only pre-image of $\underline 0$ is $\underline 0$.
Therefore, $\lambda_i = 0$, for $1 \leq i \leq r$, since the first $r$ columns or $B$ were linearly independent. Hence the first $r$ columns of $AB$ are also linearly independent. So, rank$(AB)\geq r$.
But, rank$(AB) \leq$ rank$(B) = r$ (a more elementary result). Therefore rank$(AB)=r$.  $\quad\square$
