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.
 A: A square non-singular (or invertible) is row equivalent to the identity matrix (with the same number of rows/columns).  Thus, $P$ and $Q$ can be written as the product of a finite number of elementary matrices.
Thus $PAQ$ can be formed from $A$ by performing row operations and column operations.  These operations preserve the rank of $A$.
A: Hint: multiplying matrices amounts to compose linear maps. In other words, suppose $A$ is an $n\times m$ matrix, i.e. a linear map $f_A:\mathbb R^m\to\mathbb R^n$. Then, to perform the product $PAQ$ is the same as to give a factorization of $f_{PAQ}$ as $$\mathbb R^m\overset{f_Q}{\longrightarrow}\mathbb R^m\overset{f_A}{\longrightarrow}\mathbb R^n\overset{f_P}{\longrightarrow}\mathbb R^n,$$ where $f_Q$ and $f_P$ are isomorphisms. Hence,
$$\textrm{rank}(PAQ):=\dim\,(\textrm{im}\,(f_P\circ f_A\circ f_Q))=\dim\,(\textrm{im}\,f_A)=:\textrm{rank}\,A.$$
A: 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$(AB)=r$.  $\quad\square$
A: Multiplying a matrix by a nonsingular matrix corresponds to composing some linear map (the possibly singular matrix) with an isomorphism (the invertible matrix). In particular an isomorphism (non-singular matrix) is essentially the identity map (i.e., the matrix is similar to the identity matrix), thus composing a linear map with an isomorphism preserves any linear structure the original map had to begin with, whether you compose from the left or the right.                                
