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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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    If this question is by itself on an assignment or exam, you would perhaps want to spend a bit more space explaining each step, but if it's an (a) out of (a)-(d) or something, this amount of proof will do just fine in my opinion. There are some minor notational quirks, however. It's supposed to be $A\textbf{x}=\textbf{0}$, not $A\textbf{x}=\textbf{b}$, and $A\bf{x}\mapsto\bf{x}$ means $\bf{x}$ is an eigenvector. It is better to describe $A:\Bbb R^n\to\Bbb R^n$ as one-to-one.2012-10-12
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    I made an edit to change the b, Thanks for noticing. Anyway, It's just a small exercise, but I'm not sure if the step `A must be invertible because the mapping is one-to-one` is accurate as it is.2012-10-12
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    I would not call the matrix $A$ one-to-one, but rather the transformation $x \mapsto Ax$ one-to-one.2012-10-12
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    @ChristopherA.Wong You are right. That was where I messed up my notation.2012-10-12
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    @EdwardStumperd Well, see what happens if the mapping is one-to-one (and onto). Then you can for each vector find a single one that is being mapped to it, and thus creating a map "going the other way". This map would be described by some matrix $B$, since it's linear. By observing that $BA\bf x = \bf x$ for any vector, we have that $BA = I_n$ and therefore, $A$ must be invertible, with inverse $B$. It is important that the linear map given by $A$ is both one-to-one and onto, but these two are the same when $A$ is a square matrix.2012-10-12
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    Thanks, that was exactly the link I was still missing!2012-10-12
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    Again, whether or not you need to go through an argument like that depends on the context the original question.2012-10-12
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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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