Prove that congruent matrices have the same rank.

Question

Can someone prove that two similar matrices have the same rank?

Thanks a lot.

Answer 1

Let $A$ be an $m\times n$ matrix.

Hint 1: If $G$ is an invertible $n\times n$ matrix, then $A$ and $AG$ have the same column space

Hint 2: If $F$ is an invertible $m\times m$ matrix, then $A$ and $FA$ have the same rank (because $F$ induces an isomorphism from the column space of $A$ to the column space of $FA$)

Answer 2

Suppose that $A$ and $B$ are congruent, so $A=S^TBS$. Notice that since $S$ is invertible, so too is $S^T$.Since $S$ is invertible we know that $Im(B)=Im(BS)$. Hence we also have the rank of $B$ and the rank of $BS$ are the same.Likewise, since $S^T$ is invertible, we know that $Ker(S^T(BS))=Ker(BS)$. Applying the rank nullity theorem and the equality between images , we therefore have $rank(A)=rank(S^TBS)=n-dim(Ker(BS))=n-dim(Ker(B))=rank(B)$

Answer 3

Use $\text{rank}(AB) \le \text{rank}(A)$, and $\text{rank}(AB) \le \text{rank}(B)$.