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I came across this simple exercise.

Suppose $A$ is $n\times n$ and the equation $A\textbf{x}=\textbf{0}$ has only the trivial solution. Explain why $A$ has $n$ pivot columns and $A$ is rowequivalent to $I_n$.

I understand that, because $A\textbf{x}=\textbf{0}$ implies $\textbf{x}=\textbf{0}$, $A$ must be invertible because $\bf{x} \mapsto A\bf{x}$ is one-to-one and if $A$ is invertible, it is obviously rowequivalent to $I_n$, but is there a more rigorous way to explain this?

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    @Edward Stumperd, since there is only trivial solution all row (column) vectors of $A$ are independent. So there are $n$ independent rows (columns). You get the rest from here.2012-10-12

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Consider $\mathcal{M}_n(\mathcal{F})$, the set of all $n \times n$ matrices, over a field $\mathcal{F}$. We define an equivalence relation $\cong$ on $\mathcal{M}_n(\mathcal{F})$ as follows: for $A,B \in \mathcal{M}_n(\mathcal{F})$, $A \cong B$ if there exists an invertible $P \in \mathcal{M}_n(\mathcal{F})$ such that $A=PB$. Then $I_n$ represents all $n \times n$ matrices of full rank, with respect to this row equivalence relation. Now, since $Ax=0 \Leftrightarrow x=0$, $A$ has full rank, and so it is inside the equivalence class of $I_n$, i.e. $ A \cong I_n$, i.e. there exists invertible $P$ such that $A = P I_n$. This latter relation simply says that $A$ is invertible.