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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.

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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, $$\textrm{rank}(PAQ):=\dim\,(\textrm{im}\,(f_P\circ f_A\circ f_Q))=\dim\,(\textrm{im}\,f_A)=:\textrm{rank}\,A.$$

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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 $A$.

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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.

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