Section 5.3 Dimensions of Loci
As with spans, I can use row-reduction of matrices to find the dimension of a locus. In particular, I will be able to determine if any of the equations are redundant. Recall that I described any affine or linear subspace of \(\RR^n\) as the locus of some finite number of linear equations: the set of points that satisfy a list of equations.
\begin{align*}
a_{11} x_1 + a_{12} x_2 + \ldots a_{1n} x_n \amp = c_1\\
a_{21} x_1 + a_{22} x_2 + \ldots a_{2n} x_n \amp = c_2\\
\ldots \amp\\
a_{m1} x_1 + a_{m2} x_2 + \ldots a_{mn} x_n \amp = c_m
\end{align*}
This is just a system of linear equations, which has a corresponding matrix.
\begin{equation*}
\left( \begin{array}{cccc|c}
a_{11} \amp a_{12} \amp \ldots \amp a_{1n} \amp c_1 \\
a_{21} \amp a_{22} \amp \ldots \amp a_{2n} \amp c_2 \\
\vdots \amp \vdots \amp \ldots \amp \vdots \amp \vdots \\
a_{11} \amp a_{12} \amp \ldots \amp a_{1n} \amp c_1
\end{array} \right)
\end{equation*}
So, by definition, the locus of a set of linear equations is just the geometric version of the solution space of the system of linear equations. Loci and solution spaces are exactly the same things; only loci are geometry, and solution spaces are algebra. The dimension of a locus is the same dimension as a solution space of a system. In
Section 4.3, the dimension of a solution space was determined by the free variables in the reduced form of the matrix. Let me restate the result.
Proposition 5.3.1.
Consider a locus defined by a set of linear equations. Then the dimension of the locus is the number of free variables in the solution space of the matrix corresponding to the system of the equations. If the matrix contains a row that leads to a contradiction \(0=1\text{,}\) then the locus is empty (and the dimension is not defined).
To understand a locus, I just solve the associated system. If it has solutions, I count the free variables: that’s the dimension of the locus. I can get even more information from this process. When I row reduce the matrix \(A\) corresponding to the linear system, the equations corresponding to rows with leading ones (keeping track of exchanging rows, if necessary) are equations that are necessary to define the locus. Those that end up as rows without leading ones are redundant, and the locus is unchanged if those equations are removed.
If the ambient space is \(\RR^n\text{,}\) then the equations have \(n\) variables. That is, there are \(n\) columns to the left of the vertical line in the extended matrix. If the rank of \(A\) is \(k\text{,}\) and there are no rows that reduce to \(0=1\text{,}\) then there will be \(n-k\) columns without leading ones, so \(n-k\) free variables. The dimension of the locus is the ambient dimension \(n\) minus the rank of the matrix \(k\text{.}\)
This fits the intuition for spans and loci. Spans are built up: their dimension is equal to the rank of an associated matrix since each leading one corresponds to a unique direction in the span, adding to the dimension by one. Loci are restrictions down from the total space: their dimension is the ambient dimension minus the rank of an associated matrix, since each leading one corresponds to a real restriction on the locus, dropping the dimensions by one. In either case, the question of calculating dimension boils down to the rank of an associated matrix.
Example 5.3.2.
I’m going to repeat
Example 4.3.7 to make the point that nothing new is happening here: I’m just treating loci as linear systems and looking at their solutions spaces. Consider a locus in
\(\RR^4\) defined by two equations.
\begin{align*}
w + 3y + 2z \amp = -3 \\
x - y - z \amp = -2
\end{align*}
The translation of this into a matrix is precisely the matrix in
Example 4.3.7.
\begin{equation*}
\left(\begin{array}{cccc|c}
1 \amp 0 \amp 3 \amp 2 \amp -3 \\
0 \amp 1 \amp -1 \amp -1 \amp -2
\end{array} \right)
\end{equation*}
The equation is already in reduced row-echolon form. There are two columns (other than the constant column) without leading ones; therefore \(y\) and \(z\) are free variables. The solution, in a description with free parameters, is
\begin{equation*}
\begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} =
\begin{pmatrix} -3 \\ -2 \\ 0 \\ 0 \end{pmatrix} + y
\begin{pmatrix} -3 \\ 1 \\ 1 \\ 0 \end{pmatrix} + z
\begin{pmatrix} -2 \\ 1 \\ 0 \\ 1 \end{pmatrix}\text{.}
\end{equation*}
This is an affine subspace with dimension \(2\text{.}\) The number of variables (\(4\)) minus the rank of the matrix (\(2\)) also confirms this dimension.
