Rank and Nullity is preserved by multiplication of invertible matrices (PID).

Question

I want to show that the rank and nullity of a matrix $A$ whose entries come from a PID are preserved by when $A$ is multiplied by invertible matrices

i.e If $A=PBQ$, where $P,Q$ are invertible, rank($A$)=rank($B$), nullity($A$)=nullity($B$).

Does anyone know how to show this? I feel like the lack of inverse elements is an issue. Secondly, does this condition hold for other rings in general? i.e commutative rings?

Answer 1

This is a possible answer:

Let the rank of a matrix be defined to be the maximal number of independent columns of the matrix. This is also equal to the rank of the column space of the matrix.

Refer to: Definition of rank of matrices and nullity of matrices

Suppose $A=PB$ where $P$ is invertible. Suppose columns $\{a_{i_1},...,a_{i_n}\}$ are linearly independent in $A$. Therefore if $$r_1b_{i_1}+...+r_nb_{i_n}=0,$$ then
$$r_1Pb_{i_1}+...+r_nPb_{i_n}=0,$$ which implies that $$r_1a_{i_1}+...+r_na_{i_n}=0.$$ Hence, columns $b_{i_1},..,b_{i_n}$ are linearly independent. Since we can make the same argument for $P^{-1}A=B$, we can deduce that the rank of $A$ is equal to the rank of $B$.

Suppose $A=BQ$ where $Q$ is invertible. This implies that the column space of $A$ is a subset of the column space of $B$ as the columns of $A$ are in the column space of $B$. Since $B=AQ^{-1}$, we can deduce that the $A$ and $B$ have the same column space which implies that they have the same rank.

If $A$ and $B$ have the same dimensions, by the rank-nullity theorem (see link above), the nullity of $A$ equals the nullity of $B$.