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How does the A (mxn) matrix map the row space into the column space if there is any relationship? In Ax=b, do the vectors of x have to be in the row space in order to get the solution b? b must be in the column space, right? The reason why I think that the vectors in x must be in the row space is because x is in n space and the row space is also in the n space. The column space is m sized and so is b. So if A is the linear transformation from n space to m space, then x must be valid in the row space (as a linear combination of the row space basis vectors) in order to obtain the b vector that is valid in the column space (also equivalent to a linear combination of the column space basis vectors). Is this valid thinking?

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  • $\begingroup$ $x$ need not be in the row space. Simple example: $A = 0$ and $b=0$. Then $x$ can be anything, but the row space is just $\{0\}$. $\endgroup$ – amsmath Oct 15 '19 at 3:49
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    $\begingroup$ The row space and column space have the same dimension. Apart from that there is no relation between them. $\endgroup$ – Lord Shark the Unknown Oct 15 '19 at 3:59
  • $\begingroup$ The collection of all the vectors $c$ of the form $c=Ax$ is the column space $C$. The collection of all the vectors $r$ of the form $r=yA$ is the row space $R$. They aren't connected. At least not in a very natural way. I haven't checked all the details, but I think that $(r,c)\mapsto yAx$ gives a well-defined non-degenerate bilinear pairing that allows us to somewhat naturally identify $R$ with the dual of $C$ (and vice versa). $\endgroup$ – Jyrki Lahtonen Oct 15 '19 at 4:07
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It is true that, $Ax = b$ has a solution if and only if $b \in \operatorname{colspace}A$. The reason is, if column vectors $v_1, \ldots, v_n$ are column vectors in $\Bbb{R}^{m \times 1}$, then $$\left(\begin{array}{c|c}&&&\\v_1&v_2&\cdots&v_n\\&&&\end{array}\right)\begin{pmatrix}x_1\\x_2\\\vdots\\x_n\end{pmatrix}=x_1v_1 + x_2 v_2 + \ldots+x_nv_n.$$ That is, multiplying a matrix on the right by a column vector corresponds to a linear combination of the columns. So, $Ax$ must always lie in the columnspace of $A$. Conversely, if $b$ lies in the columnspace of $A$, then the above equality provides a way of turning the linear combination into a vector $x$.

On the other hand, $Ax = b$ does not require $x$ to be in the row space of $A$. For example, take $A$ to be the $0$ matrix; you can multiply any appropriately sized column vector to the $0$ matrix, despite its rowspace being trivial. Indeed, you can always multiply an $n \times 1$ vector to an $m \times n$ matrix.

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