While reading about linear algebra for math olympiads in these notes, I came across the following assertion:
Remark. The set of invertible matrices form a Zariski (dense) open subset, and hence to verify a polynomial identity, it suffices to verify it on this dense subset.
Could someone provide an explanation of what it means to be a "Zariski (dense) open subset"? A proof of this result is sketched in the notes, but I feel there is some deeper theory going on underneath.
In case anyone is interested, the author has a similar set of notes here.
Best Answer
Zariski density means that any polynomial identity in the entries of an $n\times n$ matrix which holds on all invertible matrices holds on all matrices.
If we furthermore are considering matrices with entries in $\mathbb R$ of $\mathbb C$, then we can also say that any identity between continuous functions on the space of all $n\times n$ matrices which holds on all invertible matrices holds on all matrices.
The proof for polynomial functions is not hard:
We can work over any infinite field $k$, which we can take to be $\mathbb R$ of $\mathbb C$ if you like.
A polynomial in the entries of an $n\times n$ matrix is just a polynomial in $n^2$ variables.
Check that for any non-zero polynomial in $n^2$ variables, there is at least one matrix on which it doesn't vanish. (This is where we use that $k$ is infinite, and it ultimately reduces to the fact that polynomials in one variable have only finitely many zeroes.)
The determinant, which I'll denote $\Delta$, is a non-zero polynomial in $n^2$-variable.
Suppose that $f$ is a polynomial which vanishes on all invertible matrices. Then the product $f \Delta$ vanishes on all matrices. By the first point, it must be the zero polynomial. Since $\Delta$ is non-zero, we see that $f$ must be the zero polynomial. That is, $f$ vanishes on all matrices.
The proof for continuous functions is similar, but involves some topology as well algebra: you have to check that any non-empty open subset of $n\times n$ matrices contains an invertible matrix. This is standard, but may not be clear to you if you're not used to making arguments in topology or manifold theory.
Added: Actually, Georges's comment below gives a nice proof of the statement in the preceding comment. The same argument can also be found in Pete Clark's answer here. (This is an answer to the question linked to by Jonas Meyer a comment above.)