Is the Number of Free Variables Constant in Echelon Form Systems?

[Rasalhague]

Theorem 1.5 says that we must get the same solution set no matter how we proceed but if we do Gauss’ method in two ways must we get the same number of free variables in each echelon form system? Must those be the same variables, that is, is solving a problem one way to get y and w free and solving it another way to get y and z free impossible? In the rest of this chapter we will answer these questions. The answer to each is 'yes'.

- Jim Hefferon: Linear algebra

Theorem 1.5 (Gauss' method). If a linear system is changed to another by one of these operations

(1) an equation is swapped with another
(2) an equation has both sides multiplied by a nonzero constant
(3) an equation is replaced by the sum of itself and a multiple of another

then the two systems have the same set of solutions.

Here is a counter example to the claim that the number of free variables is constant. Take the following coefficient matrix in echelon form:

\begin{bmatrix}<br /> 2 &amp; 3\\ <br /> 0 &amp; 1<br /> \end{bmatrix}

It shows one free variable, by the definition quoted below. But it can be transformed by row operation 3 into to another matrix, also in echelon form, which shows no free variables:

\begin{bmatrix}<br /> 2 &amp; 0\\ <br /> 0 &amp; 1<br /> \end{bmatrix}.

1.10 Definition In each row of a system, the first variable with a nonzero coefficient is the row’s leading variable. A system is in echelon form if each leading variable is to the right of the leading variable in the row above it (except for the leading variable in the first row).

2.2 Definition In an echelon form linear system the variables that are not leading are free.

Should "echelon form" be replaced with "reduced echelon form" in definition 2.2 and the statement I quoted at the beginning of this post?

 

[Rasalhague]

Latest thoughts: Variables correspond to columns of the coefficient matrix, rather than entries. So would it be fair to say that the definition "In each row of a system, the first variable with a nonzero coefficient is the row’s leading variable." is misleading because leading variables are a property of the system, not of an individual row?

Maybe better to say, "A leading variable, x_j, of the system is one for which the corresponding column in any echelon form of the coefficient matrix is a pivot column (that is, there exists a row, row i, such that the entry A_ij has no nonzero entries preceding it in row i. If no such row exists, the variable corresponding to that variable is free."

So the number of free variables is a constant of the system, and it's customary when using Gaussian elimination to treat the free variables as parameters when describing the solution set, although leading variables could in principle be used. The choice of which variables to label as leading and which free is due to certain conventions in the method of Gaussian elimination (e.g. the fact that an upper triangular matrix is used), rather than being an inherent property of these variables (inherent to the structure of the linear system).

 

[chiro]

Remember that for a linear system, we are usually dealing with Ax = b for matrix A, vector b and unknown x. Gaussian operations preserve the system because changes on A will induce changes in b and make the equality condition hold.

 

[isaac200]

That's very good