# 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?

• Well it depends on what your definition of a plane is. Some people call any k dimensional linear subspace of $R^n$ a (higher dimensional) plane if $k\ge 2$, which is apparently what your source did. Then for some other people a plane has to be 2-dimensional, and higher dimensional analogues are called flats. In either case the solution space in your example is a plane in $R^4$. – Hyperplane Mar 15 '17 at 19:23
• How fitting your username is...on the good ole Wiki I see that "For $n$ variables, each linear equation determines a hyperplane in $n$-dimensional space. The solution set is the intersection of these hyperplanes, which may be a flat of any dimension." Can't wrap my head around it too much, but I think that answers my question (wish the author had made an aside, since he actually does talk about hyperplanes near the end of the text). Thanks for the clarifying comment. – interrogative Mar 15 '17 at 19:30

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 $6$.
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.).