Basic question on the infinitely many solutions of a linear system Ax=b, I just want to verify the geometry of solutions to $Ax=b$, for the case when we have infinitely many solutions:
If say for a $3\times 3$ matrix, after Gaussian Elimination, I have two pivot variables and one free variable - for simplicity, assume the free variable is the third coordinate $x_3$.
Then the infinitely many solutions form a line in $\mathbb{R}^3$.
Is there a similar geometric intuition for when I have only one pivot variable, and two free variables?  Do the infinitely many solutions form a plane in $\mathbb{R}^3$?  I am pretty sure it does, but just wanted to ask the question, in case I am mistaken.  The solutions would be in the form of linear combinations of two linearly independent vectors, so I think these two vectors will span a plane.  
Thanks,
 A: If you have one pivot variable, your reduced (augmented) matrix is of the form
$$\left[\begin{array}{ccc|c}a&b&c&d\\0&0&0&0\\0&0&0&0\end{array}\right]$$
then the solution is of the form $ax+by+cz = d$, which is the equation of a plane.
A: More generally, if $n$ is the nullity of the matrix corresponding to a homogeneous linear system, then the solution space is a $n$-dimensional subspace (it passes through the origin). If we have a corresponding consistent non-homogeneous system, the solution space is the affine space obtained from the aforementioned subspace by a translation through any particular solution.
A: A way to think about this is, we let $x=x_p+x_h$, so that $x_p$ is the particular solution and $x_h$ is the homogeneous solution.
We can fix $x_p$ and for any other solution $x$ to the equation, $x_h=x-x_p$ is in the nullspace of $A$.
$Ax=b, Ax_p=b\Rightarrow A(x-x_p)=b-b=0\Rightarrow Ax_h=0$
Since the nullspace is a vector space, adding $x_p$ back to it is "shifting" it to the correct location.
Hence all solution sets are "shifted" vector spaces or the empty set, in case of no solution.
