Are a square matrix's columns and rows either both(separately) linearly independent or both(separately) linearly dependent? Prove or disprove:
Given a square matrix $A$,the columns of $A$ are linearly independent iff. the rows of $A$ are linearly independent.
 A: A more intuitive argument
(why for a square (nxn) matrix ,if it's rows are linearly dependent it implies that so are also it's columns)
Given that rows are linearly dependent,performing Gauss elimination must produce  row of all zeroes (possibly more than one)  .
Therefore the  columns of the row reduced echelon form  matrix are linearly dependent.
That's so because all have zero in the same entry,thus have only n-1 "free" variables ,non-zero entries ,thus are n-1 dimensional,and n vectors in only n-1 dimensional space- can't  be linearly independent.
And since columns of the reduced matrix are linearly depependent , so must  be also the columns of the original matrix,which is our claim.
That's so because  Gauss elimination  preserves relation between columns!
More specifically if some column of the  original matrix is some  linear combination of other columns, then the corresponding columns  of the   row reduced ,echelon form,matrix mirror this precisely.  (otherwise it wouldn't be possible ,for example ,to find  null space of the  original matrix ,by examining the much simpler columns of it's row reduced echelon form)
A: 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. 
