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

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?

• if P and Q are invertible, they are full rank. Commented Nov 5, 2021 at 16:43
• How do you define the rank of a matrix over a PID?
– user700480
Commented Nov 6, 2021 at 12:20

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 $$B$$.