Converting to standard form - at basic solutions, at least one of the two positive variables that replaces a free variable should be zero. Why?

Question

My linear optimization note says the following about handling free variables (variables that are not restricted to be non-negative) when converting a linear problem into the standard form:

The method to deal with free variables is to replace them with the difference of two variables $(x^+ - x^-)$, both non-negative. At basic solutions, at least one of the two variables takes the value zero.

I don't see why this happens. I do know that, if you have $n$ variables and $m$ equality equality constraints then at basic solutions $(n-m)$ of the variables must take the value zero. But I do not understand why one of $x^+$ and $x^-$ has to be zero.

Is there any proof for this? Any explanation would be appreciated.

@LinAlg sorry, it should be non-negative. These are notes I took down myself when my prof, lectures. There could be errors. Let me correct it. — user5393678, Nov 01 '18 at 20:59

score 1 · Accepted Answer · answered Nov 01 '18 at 21:06

1

If one of the variables is in the basis, its corresponding column is $e_i$, and the column of the other variable is $-e_i$. The other variable therefore cannot get selected for entering the basis (its reduced cost is 0).

Another reason why they cannot both be in the basis is because then the columns of the basis would be linearly dependent.

answered Nov 01 '18 at 21:06

LinAlg

20,093

Is there any rigorous proof you can provide? – holala Oct 01 '20 at 16:18
@holala can you indicate why the linear dependence argument is not rigorous? – LinAlg Oct 01 '20 at 18:24

Converting to standard form - at basic solutions, at least one of the two positive variables that replaces a free variable should be zero. Why?

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