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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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    @POTUS What you are trying to prove is almost the definition of the column space of a matrix.2012-07-02

0 Answers 0