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$A$ 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 $M$A:\Bbb R^N\to \Bbb R^M$ must have nontrivial kernel... – 2012-10-10
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    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 $im(A) = \mathbb R^m$.2012-10-10
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    $\rank(A)=$ the number of linearly independent columns, and $\rank(A)=$the number of linearly independent rows.......2012-10-10

3 Answers 3

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

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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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    @Justin If you liked this answer or any of the other answers provided by the other users then you can click the check mark beside the answer to "accept" it. If you have 15 rep or above, then you can even upvote an answer to show gratitude; you can upvote multiple answers. This will help the user and yourself.2013-11-23