The solution set of a system of $n$ linear equations written as a matrix equation $AX=b$,$b\neq0$ cannot have zero vector in it. This means that the solution set cannot be a vector space and talking about it's dimensions is absurd.
But why do then people talk about the dimension of the solution set as in here :.
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$\begingroup$The set of all solutions for $Ax = b$ is only a vector space iff $b=0$, as you said, and if that is the case you will have your normal definition of dimension.
But if $b \neq 0$, the set of all solutions will be an Affine space. An affine space is basically a point plus a vector space.
For example, let $V$ be the set of solutions for $Ax = 0$, and let $k$ be just one solution for $Ax = b$. Then, the set of ALL solutions of $Ax = B$ will be: $$\{k\} + V$$
And because this is a point plus a vector space this is an Affine space. By definition, the dimension of an Affine space $\{k\} + V$ is just the dimension of the vector space $V$ and that is why even if the space of all solutions for $Ax = B$ isn't a vector space, you can still define its dimension.
$\endgroup$ 6 $\begingroup$That is true. In fact, if $b \neq 0$, then the set of solutions of the system $Ax = b$ is not a vector space (because $0$ will not be an element as you well pointed out). Although not being a vector space, that set will constitute an affine space. And the dimension of this affine space, will be the dimension of the vector space generated by the set of the solutions of the homogenous system associated $Ax = 0$.
$\endgroup$ 0 $\begingroup$If $A$ is a $m \times n$ matrix and if $b \in \mathbb R^m,$ then the soltion set $L$ of $Ax=b$ has the form
$$ L=x_0 +U,$$
where $U$ is a subspace of $ \mathbb R^n$ and $x_0$ is a special solution of $Ax=b.$ Then
$$ \dim L := \dim U.$$
Note that $U$ is the solution space of $Ax=0.$
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