# 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.

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i.e. "$\mathbf A$ is singular iff $\mathbf A^\top$ is singular"... – J. M. Jul 20 '12 at 7:23
Column rank = row rank or $rank(A) = rank(A^T)$ . Read this – sabertooth Jul 20 '12 at 7:26
2 answers, none of them "acceptable"... – draks ... Jul 20 '12 at 7:33
@J.M. Linearly independent rows of $A$ are linearly independent columns of $A^T$, and linearly independent columns of $A^T$ make $A^T$ invertible, which in turn makes $A$ invertible, which finally gives linearly independent columns of $A$. The reverse is also true. Hence, we have proved the problem statement. However, I'm having difficulty accepting/"seeing" the proof intuitively, even though I can logically make the connections. Help! – Ryan Jul 20 '12 at 7:37
I didn't answer, simply because you haven't mentioned what you're allowed to use. For instance, one (lazy) way of seeing this is that since a matrix and its transpose are similar matrices, then the singularity of one implies the singularity of the other. But that might be too high-powered for the matter at hand... – J. M. Jul 20 '12 at 7:50

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.

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I understand the proof, but still can't quite believe that columns of a square matrix being LI $\implies$ its rows are LI too. Perhaps I'll get it in the shower some day. – Ryan Jul 20 '12 at 22:28
If you believe rows of $A$ lin. indep. implies columns of $A$ lin. indep., then you believe rows of $A$-transpose lin. indep. implies columns of $A$-transpose lin. indep. But the rows of $A$-transpose are the columns of $A$, and the columns of $A$-transpose are the rows of $A$. – Gerry Myerson Jul 21 '12 at 5:29
I can't see how this answers the question. – Don Larynx May 10 '15 at 3:12
@Don, I'm sorry to hear that. But if you want someone to help you see how it answers the question, you are going to have to be a lot more forthcoming about just what it is that you don't see. – Gerry Myerson May 10 '15 at 6:10
I didn't know about this definition: "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." Where did you read it? – Don Larynx May 10 '15 at 14:32