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Course Notes for Linear Algebra

Remkes Kooistra

Section 4.3 Description by Parameters

Subsection 4.3.1 Solution Spaces and Free Parameters

Definition 4.3.1.

If I have a system of equations in the variables \(x_1, x_2, \ldots, x_n\text{,}\) the set of values for these variables which satisfy all the equations is called the solution space of the system. Since each set of values is a vector, I think of the solution space as a subset of \(\RR^n\text{.}\) For linear equations, the solution space will always be an affine subspace. If I have encoded the system of equations in an extended matrix, we can refer to the solution space of the matrix instead of the system.
The term ‘solution space’ is a conventional term in algebra, since it refers to the solution of some equations. However, geometrically, I haven’t actually done anything knew. The set of points that satisfy some number of equation is the locus of those equations. So solutions spaces are just another name for loci. The description of solutions spaces in this section can also be understood as descriptions of loci.

Definition 4.3.2.

Since solution spaces are affine, they can be written as \(u + L\) where \(u\) is a fixed offset vector, and \(L\) is a linear space. Since \(L\) is linear, it has a basis \(v_1, v_2, \ldots v_k\text{,}\) and any vector in \(L\) is a linear combination of the \(v_i\text{.}\) I can write any and all solutions to the system in the following fashion (where \(a_i \in \RR\)).
\begin{equation*} u + a_1v_1 + a_2v_2 + a_3v_3 + \ldots + a_kv_k \end{equation*}
In such a description of the solution space, I call the \(a_i\) free parameters. One of the goals in solving systems is to determine the number of free parameters for each solution space.
I want to show how to go from reduced row echelon matrices to descriptions of solutions spaces by parameters. This is best to do with a series of examples showing all the various posibilities.

Example 4.3.3.

Let me start with a matrix that encodes a unique solution.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 8 \end{array} \right) \end{equation*}
This example is the easiest case: there is a leading one in each row and column left of the horizontal line, and I just read off the solution, \(x=-3\text{,}\) \(y=-2\) and \(z=8\text{.}\) There are no parameters, since the solution is unique. The ‘description by parameters’ is just the statement of the unique solution.

Example 4.3.4.

I might have a row of zeros in the reduced row-echelon form.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 8 \\ 0 \amp 0 \amp 0 \amp 0 \end{array} \right) \end{equation*}
This row of zeros doesn’t actually change the solution at all: it is still \(x=-3\text{,}\) \(y=-2\) and \(z=8\text{.}\) The last line just corresponds to an equation that says \(0x + 0y + 0z = 0\) or just \(0=0\text{,}\) which is always true. Rows of zeros arise from redundant equations in the original system. This still have a unique solution and no free parameters.

Example 4.3.5.

There might be a column of zeros. Consider something similar to the previous examples, but now in four variables \(w,x,y,z\text{.}\) (In that order, so that the first column is the \(w\) column, the second is the \(x\) column, and so on).
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp 0 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 0 \amp 8 \end{array} \right) \end{equation*}
The translation gives \(w=-3\text{,}\) \(x=-2\) and \(y=8\text{,}\) but the matrix says nothing about \(z\text{.}\) A column of zeros corresponds to a free variable: any value of \(z\) solves the system as long as the other variables are set. This system has one free paramter. A description of the solution by parameters here would show that the first three coefficients are constant, but \(z\) can be anything. I can express this as a sum of vectors.
\begin{equation*} \begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \\ 8 \\ 0 \end{pmatrix} + z \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix} \end{equation*}

Example 4.3.6.

The columns of free variables need not contain all zeros. The following matrix is still in reduced row-echelon form.
\begin{equation*} \left( \begin{array}{cccc|c} 1 \amp 0 \amp 0 \amp 2 \amp -3 \\ 0 \amp 1 \amp 0 \amp -1 \amp -2 \\ 0 \amp 0 \amp 1 \amp -1 \amp 8 \end{array} \right) \end{equation*}
Reduced row-echelon form needs a leading one in each non-zero row, and all other entries in a column with a lead one must be zero. This matrix satisfies all the conditions. The fourth column corresponds to the \(z\) variable but lacks a leading one. Any column lacking a leading one corresponds to a free variable. Look at the translation of this matrix back into a linear system.
\begin{align*} w + 2z \amp = -3\\ x - z \amp = -2\\ y - z \amp = 8 \end{align*}
I move the \(z\) terms to the right side.
\begin{align*} w \amp = - 2z - 3\\ x \amp = z - 2\\ y \amp = z + 8\\ z \amp = z \end{align*}
I have included the last equation, which is trivial to satisfy, to show how all the terms depend on \(z\text{.}\) This makes it clear that \(z\) is a free parameter. I can write the solution as follows in terms of that free parameter.
\begin{equation*} \begin{pmatrix} w \\ x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -3 \\ -2 \\ 8 \\ 0 \end{pmatrix} + z \begin{pmatrix} -2 \\ 1 \\ 1 \\ 1 \end{pmatrix} \end{equation*}
Any choice of \(z\) will solve this system. If I want a specific solution, I take specific values of \(z\text{.}\) For example, if \(z=1\text{,}\) these equations give \(w=-5, x=-1\) and \(y=9\text{.}\)
Moreover, this description by parameters describes the solutions as an offset span. The columns of constants is the offset and the free variable gives the span, in this case, just the span of one vector.

Example 4.3.7.

The following matrix is also in reduced row-echelon form (again, using \(w\text{,}\) \(x\text{,}\) \(y\) and \(z\) as variables in that order).
\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*}
Here, both the third and the fourth columns have no leading one; therefore, both \(y\) and \(z\) are free variables. I translate the matrix back into a linear system.
\begin{align*} w \amp = -3y - 2z - 3\\ x \amp = y+z-2\\ y \amp = y\\ z \amp = z \end{align*}
Then I can write the solution using the free parameters. I can express the four equations as one equation of vectors; each equation is the equation of one of the four components of the vectors, filling in zeros for anything not mentioned in the equations.
\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} \end{equation*}
Any choice of \(y\) and \(z\) gives a solution. For example, if \(y=0\) and \(z=1\) then \(w=-5\) and \(x=-1\) complete a solution.

Example 4.3.8.

I can also have a reduced row-echelon matrix of the following form.
\begin{equation*} \left( \begin{array}{ccc|c} 1 \amp 0 \amp 0 \amp -3 \\ 0 \amp 1 \amp 0 \amp -2 \\ 0 \amp 0 \amp 1 \amp 8 \\ 0 \amp 0 \amp 0 \amp 1 \end{array} \right) \end{equation*}
The first three rows translate into equations without issue. However, the fourth row translates to \(0x+0y+0z = 1\) or \(0=1\text{.}\) Obviously, this can never be satisfied. Any row which translates to \(0=1\) leads to a contradiction. A contradition means that the system has no solutions, since the \(0=1\) can never be satisfied.
I will summarize the three possible situations for reduced row-echelon matrices.
  • If there is a row that translates to \(0=1\text{,}\) then there is a contradiction, and the system has no solutions. This overrides any other information: whether or not there are free variables elsewhere, there are still no solutions.
  • If there are no \(0=1\) rows and all columns left of the vertical line have leading ones; then there is a unique solution. There are no columns representing free variables, and each variable has a specific value that can be ead directly from the matrix.
  • If there are no \(0=1\) rows and there is at least one column left of the vertical line without a leading one; then each such column represents a free variable. The remaining variables can be expressed in terms of the free variables, and any choices of the free variables lead to a solution. There are infinitely many solutions. The solutions space can always be expressed as an offset span.
In particular, note that there are only three basic cases for the number of solutions: none, one, or infinitely many. No linear system has exactly two or exactly three solutions. In the case of infinitely many solutions, the interesting question is the dimension of the solution space. This question, however, is easily answered.
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