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Lets generate a random symmetric matrix $A$ by generating the entries of a random matrix $Z$ iid from some continuous distribution, and setting $A=(1/2)(Z+Z^T)$. I think its true that $A$ should have distinct eigenvalues with probability $1$, and it seems like there should be a really simple argument to this effect, but I am not seeing it.

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up vote 9 down vote accepted

A matrix $A$ has distinct eigenvalues if and only if the discriminant of its characteristic polynomial is nonzero. This is a polynomial function of the entries of $A$, so the set of points at which it is equal to zero is about as well-behaved as possible; in particular it has measure zero in any reasonable distribution.

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