Row equivalent matrices have the same row rank

Prove that two row equivalent matrices over a field have the same row rank.

Let $F$ be a field.

Suppose $A$ and $C$ are row equivalent via the matrix $P$, which is a product of elementary matrices. That is, $PA = C$, so that $A^\mathsf{T} P^\mathsf{T} = C^\mathsf{T}$. Let $B$ and $E$ denote the standard bases, and say $A^\mathsf{T} = M^E_B(\varphi)$, $C^\mathsf{T} = M^E_B(\psi)$, and $P^\mathsf{T} = M^B_B(\theta)$. In particular, we have $\varphi \circ \theta = \psi$. By this previous exercise, the row rank of $A$ is the column rank of $A^\mathsf{T}$ is the dimension of $\mathsf{im}\ \varphi$, and the row rank of $C$ is the column rank of $C^\mathsf{T}$ is the dimension of $\mathsf{im}\ \psi$. Since $\varphi \circ \theta = \psi$ and $\theta$ is an isomorphism, we have $\mathsf{im}\ \varphi \subseteq \mathsf{im}\ \psi$, and likewise since $\varphi = \psi \circ \theta^{-1}$, $\mathsf{im}\ \psi \subseteq \mathsf{im}\ \varphi$. Thus we have $\mathsf{im}\ \varphi = \mathsf{im}\ \psi$, so that $A$ and $C$ have the same row rank.

Advertisements
Post a comment or leave a trackback: Trackback URL.