# Linear Independent Rows vs. Columns

$A$ is an $M\times N$ matrix with linearly independent rows and linearly independent columns. Prove that $A$ must be square matrix.

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Assume wlog that $M<N$. Then $A:\Bbb R^N\to \Bbb R^M$ must have nontrivial kernel... –  anon Oct 10 '12 at 21:27
You also can see this by the dimensionforumula for functions in $Hom(\mathbb R^n, \mathbb R^m)$. Use this to conclude that $\dim \ker A >0$ with $im(A) = \mathbb R^m$. –  Andre Oct 10 '12 at 21:48
$\rank(A)=$ the number of linearly independent columns, and $\rank(A)=$the number of linearly independent rows....... –  N. S. Oct 10 '12 at 23:07

Hint: The matrix must have full row and column rank, but $\mathrm{rank} (A) \le \min(m,\ n)$

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Assume $M\leq N$ (otherwise we can work with $A^T$). So we have $N$ columns, each a vector in $\mathbb{R}^M$; the maximum number of linearly independent vectors in $\mathbb{R}^M$ is $M$, so we have $N\leq M$. Then $N=M$.

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If $V$ is a vector space with $\dim V=k$, and you pick $v_1,\, v_2,\ldots,\, v_p$ in $V$ with $p\gt k$ then those vectors must be l.d. because otherwise we would have a subspace $W$ of $V$ with $\dim W\gt \dim V$, namely $W=\langle v_1,\ldots,v_p\rangle$.

If $A$ is a matrix with real entries and $M\lt N$, the columns of $A$ are $N$ vectors of $\Bbb R^M$ therefore they are l.d. Since this is contradicts that $A$ have linearly independent columns, it must be $M\geq N$.

Now, $A^\text{T}$ is a $N\times M$ matrix with linearly independent rows and linearly independent columns. By what we already have, it follows that $N\geq M$.

Therefore $M=N$.

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