Prove or disprove:
Given a square matrix $A$,the columns of $A$ are linearly independent iff. the rows of $A$ are linearly independent.
Here's an argument more-or-less from first principles.
If the rows of $A$ are linearly independent, then the result of doing row-reduction to $A$ is the identity matrix, so the only solution of $Av=0$ is $v=0$.
If the columns of $A$ are linearly dependent, say, $$a_1c_1+a_2c_2+\cdots+a_nc_n=0$$ where the $c_i$ are the columns and the $a_i$ are not all zero, then $Av=0$ where $$v=(a_1,a_2,\dots,a_n)\ne0$$
So, if the columns are dependent, then so are the rows.
Now apply the same argument to the transpose of $A$ to conclude that if the rows of $A$ are dependent then so are the columns.