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Prove: If $A$ and $b$ are $m\times n$ and $m\times 1$ matrices, respectively, over a field $\mathbb{F}$, then the matrix equation $Ax=b$ is solvable over $\mathbb{F}$ if and only if $b$ is in the column space of $A$.

Here is an attempt to prove it:

($\Rightarrow $) Suppose that $Ax=b$ is solvable over $\mathbb{F}$ where the columns of $A$ are $a_1,a_2,\cdots, a_n $ . We want to show that $b\in CS(A).$ Then observe that $Ax=(a_1 a_2 \cdots a_n)x=a_1x+a_2x+\cdots +a_nx=b$, so $b$ is a linear combination of the columns of $A$. Thus $b\in $ span (Columns of $A$)=$CS(A).$

($\Leftarrow $) Suppose that $b\in CS(A)$. Then $b$ can expressed as a linear combination of the columns of $A$. So, there exists an $x\in \mathbb{F}^{n\times 1}$ such that $b=a_1x+a_2x+\cdots+a_nx=(a_1 a_2 \cdots a_n)x=Ax$. Hence $Ax=b$ is solvable over $\mathbb{F}$.

Please comment/suggest/correct it.(Note that I have removed the flawed proof written earlier.)

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Do you actually have a definition of column space that gives it $\{Ax\mid x\in\mathbb{F}^{n\times 1}\}$? Normally, the column space is defined to be the space spanned by the columns, and then one proves that $b$ is in the column space if and only if there exists $x$ such that $Ax=b$. In fact, that's what this problem is asking you to prove. So it seems to me that unless you were actually given the definition of columnspace you are using, then your argument is circular. – Arturo Magidin Jul 1 '12 at 19:12
@ArturoMagidin Yes, we defined column space of $A$, as $\{Ax|x\in \mathbb{F}^{n\times 1}\}$ or span(columns of $A$). My professor said either definition is fine when I took the class. – Lyapunov Jul 1 '12 at 19:18
Hmmm... That's an equivalent definition, of course, but I doubt it's the intended definition for this problem. Why don't you prove that the space spanned by the columns is equal to $\{Ax\mid x\in\mathbb{F}^{n\times 1}\}$? – Arturo Magidin Jul 1 '12 at 19:27
@ArturoMagidin I see what you are saying! You are right I assumed what I wanted to prove! Thanks, I will post a revised proof soon. – Lyapunov Jul 1 '12 at 19:28
@POTUS What you are trying to prove is almost the definition of the column space of a matrix. – user38268 Jul 2 '12 at 0:59

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