# 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.

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 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 $A$.

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$$(AB)=r$$. $$\quad\square$$

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.