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Free variables constant?

  1. Apr 20, 2012 #1
    - 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:

    [tex]\begin{bmatrix}
    2 & 3\\
    0 & 1
    \end{bmatrix}[/tex]

    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:

    [tex]\begin{bmatrix}
    2 & 0\\
    0 & 1
    \end{bmatrix}.[/tex]

    Should "echelon form" be replaced with "reduced echelon form" in definition 2.2 and the statement I quoted at the beginning of this post?
     
    Last edited: Apr 20, 2012
  2. jcsd
  3. Apr 21, 2012 #2
    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, xj, 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 Aij 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).
     
  4. Apr 21, 2012 #3

    chiro

    User Avatar
    Science Advisor

    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.
     
  5. Apr 25, 2012 #4
    That's very good
     
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