## Row equivalent matrices have the same column rank

Let $A$ and $B$ be row equivalent matrices over a field, say by an invertible matrix $P$ such that $PA = B$.

1. Prove that, for all matrices $X$ (such that the dimensions match), $AX = 0$ if and only if $BX = 0$.
2. Prove that any linear dependence satisfied by the columns of $A$ is also satisfied by the columns of $B$.
3. Conclude that $A$ and $B$ have the same column rank.

1. If $AX = 0$, then $PAX = P0$, so $BX = 0$. Conversely, if $BX = 0$, then $P^{-1}BX = P^{-1}0$, so $AX = 0$.
2. Write $A$ and $B$ as column matrices: $A = [A_1 | \cdots | A_n]$ and $B = [B_1 | \cdots | B_n]$. Note that, since $PA = B$, we have $[PA_1 | \cdots | PA_n] = [B_1 | \cdots | B_n]$. In particular, $PA_i = B_i$ for all $i$. Now suppose $\sum \alpha_i A_i = 0$ is a linear dependence among the columns of $A$. Now apply $P$ (as a linear transformation) to this equation; we have $0 = P(\sum \alpha_i A_i) = \sum \alpha_i PA_i = \sum \alpha_i B_i$. Thus the columns of $B$ satisfy the same linear dependence.
3. Suppose $\{A_i\}_K$ is a maximal linearly independent set of columns of $A$. By part (2), $\{PA_i\}_K = \{B_i\}_K$ is a linearly independent set of columns of $B$. If this set is not maximal, then by part (2) again (using $P^{-1}$) $\{A_i\}_K$ is not maximal, a contradiction. So $\{B_i\}_K$ is a maximal linearly independent set. Since $P$ is injective (as a linear transformation), these sets have the same cardinality. In particular, $A$ and $B$ have the same column rank.