Visualizing the solution set of $A\mathbf{x}=\mathbf{0}$.

Question

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?

Answer 1

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 $6$.

Answer 2

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.).