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This is a continuation of a previous question .

Define that elements of $x=(x_1,x_2,...,x_n)$ are distinct if for each $i≠j, x_i≠x_j.$ Consider a system of linear equation $Ax=b$. We want to understand the property of $(A,b)$ such that the solution $x$ given $A$ and $b$ is distinct in the sense defined above.

In a previous question, Theo kindly showed that the set of such $(A,b)$ is open and dense. Now we want to extend the result in a probabilistic (or measure-theoretic) sense. For example,

Suppose we pick A and b randomly. (For example, suppose that elements of A and b are chosen according to independent uniform distribution on [0,1].) Then calculate x when possible. Now, can it be that x has distinct elements in the sense defined in (a) with probability 1?

I also want to understand whether Sard's theorem can be applied to this setting. I will appreciate your comments.

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Yes, the probability of getting x with all components distinct is 1.

The question about probability is really a matter of measure theory. The probability that matrix A is nonsingular is 1, and for fixed matrix A it is clear that the set of b such that two or more components of $x = A^{-1} b$ are equal has measure zero (since it can be covered by a union of finitely many sets of measure zero, each specifying equality of exact two components of x).

Combining this with the fact that the singular matrices A have measure zero, we get the desire result.

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    @Thales: For a fixed "smooth" system with $n$ unknowns and $n$ equations, if the [Jacobian](http://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant) is bounded away from zero, then an argument can be made that the preimage of the union of hyperplanes having pairs of equal coordinates again has measure zero. Advancing the argument to cover a "space" of smooth and differentiable functions would be complicated if a uniform bound away from zero of these determinants is lacking. In the linear case the argument is simplified by the determinant being a constant (for fixed A).2011-03-28