Visualizing the solution set of $A\mathbf{x}=\mathbf{0}$. For the solution set of the homogeneous equation $A\mathbf{x}=\mathbf{0}$, I am told I can visualize it as follows:


*

*the single point $\mathbf{0}$, when $A\mathbf{x}=\mathbf{0}$ has only the trivial solution,

*a line through $\mathbf{0}$, when $A\mathbf{x}=\mathbf{0}$ has one free variable,

*a plane through $\mathbf{0}$, when $A\mathbf{x}=\mathbf{0}$ has two free variables. (For more than two free variables, also use a plane through $\mathbf{0}$.)


My question really concerns the last point. I feel as though I understand everything except for how more than two free variables corresponds to a plane through $\mathbf{0}$.
In general, using two variables, we have a line. Using three variables, we have a plane. Using four or more variables...what do we have then? For instance, the homogeneous system 
$$
x_1-3x_2-9x_3+5x_4=0\\[0.5em]
x_2+2x_3-x_4=0
$$
row reduces as
$$
\begin{bmatrix}
1 & -3 & -9 & 5 & 0\\
0 & 1 & 2 & -1 & 0
\end{bmatrix}
\sim
\begin{bmatrix}
1 & 0 & -3 & 2 & 0\\
0 & 1 & 2 & -1 & 0
\end{bmatrix}
$$
to yield the solution set
$$
\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}
=
\begin{bmatrix}
3x_3-3x_4\\
-2x_3+x_4\\
x_3\\
x_4
\end{bmatrix}
=x_3
\begin{bmatrix}
3\\
-2\\
1\\
0
\end{bmatrix}
+x_4
\begin{bmatrix}
-2\\
1\\
0\\
1
\end{bmatrix}.
$$
Apparently an "appropriate geometric picture" for the solution set is a plane through the origin.
Can someone explain this? Why can the solution set be viewed as a plane through the origin? Why can one generally look at the solution set as a plane through the origin when there are two or more free variables? Is this entirely accurate?
 A: A plane in a finite dimensional vector space can be described as a two dimensional subspace.
Note that,  when a subspace $F$ is defined by a set of linear equations (after a basis has been chosen) in a vector space $E$ of dimension $n$, the codimension of the subspace, i.e. the number
$$\operatorname{codim}_EF=\dim E-\dim F,$$
is equal to the rank of the system of equations, i.e. the rank of its associated  matrix.
If the codimension is, say, $3$ (which means the subspace of solutions is defined by $3$ linearly independent equations),  we have a subspace of dimension $n-3$: it will be a line if the  big space $E$ has dimension $4$, a plane if it has dimension $5$, a ‘solid’ (isomorphic to $K^3$) if it has dimension $6$.
A: If you think that visualizing $\mathbb R^n$ is not difficult, then it is also not difficult to see that the solution set
$$\ker A:=\{\vec{x} \in \mathbb R^n \mid A\vec{x}=0\}$$
is a $k$ dimensional vector space, where $k=n-\mathrm{rank} A$. But since it is a linear subspace, it has a basis $e_1,..,e_k$. Send each of these to the standard basis in $\mathbb R^k$ and observe that we obtain an isomorphism. Moreover, if the basis you specified originally was orthonormal, we have an isometric isomorphism, so in a rigorous sense, everything is the same (including inner product etc.).
