# Invertible Matrices are dense

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.

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The determinant is a polynomial $P$ with respect to the entries, and invertible matrices are the matrices $A$ such that $P(A)\neq 0$ (the set $\{P=0\}$ is Zariski closed since it's the zeros of a polynomial, Zariski closed are the common zeros of a family of polynomials). –  Davide Giraudo Jun 2 '12 at 20:34
How does knowing that show that it's "dense" in the sense that verifying something just for the invertible matrices verifies it for all of them? –  Potato Jun 2 '12 at 20:35
The deeper theory is, I guess, all of algebraic geometry. –  Dylan Moreland Jun 2 '12 at 20:41
What do you know about point-set topology? –  Qiaochu Yuan Jun 2 '12 at 21:19
A related question: math.stackexchange.com/questions/18964/… –  Jonas Meyer Jun 3 '12 at 1:11

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.)

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Nice answer, Matt. Just for fun, let me take up your implicit challenge to find an elementary reason why a non-empty open subset $U$ of all matrices contains an invertible matrix: Take $A\in U$ at random and almost all $A-\varepsilon Id$ will be invertible because $A$ has only finitely many eigenvalues. Friendly greetings –  Georges Elencwajg Jun 3 '12 at 16:17
@GeorgesElencwajg: Dear Georges, Nicely done! Best wishes, –  Matt E Jun 3 '12 at 17:22

A Zariski topology is a particular topology defined on $\mathbb{R}^n$ (or in more general space but it is not important here). Let $\mathbb{R}[X_1,\dots,X_n]$ be the ring of polinomials in $n$ variable. The closed sets of the Zariski topology of $\mathbb{R}^n$ are of the form $V(S) = \{a \in \mathbb{R}^n\mid \forall f\in S,\ f(a) = 0\}$ where $S$ is a subset of $\mathbb{R}[X_1,\dots,X_n]$.

The set of square matrices of dimension $n$ are, as a vector space, isomorphic to $\mathbb{R}^{n^2}$. Now, we can equip $\mathcal{M}(\mathbb{R}, n)$ with the Zariski topology using this isomorphism.

The determinat can be seen as a polynomial in $n^2$ variables (one for any the matrix entries) so the set of the non-invertible matrices is a closed set in the Zariski topology. And obviously the set of the invertible matrices are an open set.

The Zariski topology is studied a lot by algebraic geometer and, from a topological point of view, is a very bad topology. It is not hausdorff for example.

Anyway it has a very strange characteristic: it is an irreducible space. That means that:

1. No two open set are disjoint
2. the space cannot be written as the union of two closed sets.
3. every nonempty open set is dense
4. the interior of every closed set in empty
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1. non-empty open subsets; 2. proper closed subsets; 4. proper closed subsets. One can also notice that these four properties are equivalent. –  user18119 Jun 2 '12 at 21:00
How does this lead to the result in the remark (about verifying identities only for invertible matrices)? –  Potato Jun 2 '12 at 21:08
Yes, they are. It's, indeed, a quite easy topological exercise. However It's not equally immediate that the Zariski topology is irreducible. –  Vittorio Patriarca Jun 2 '12 at 21:10
Because if the polynomial identity is valid for every invertible matrix, then you have a closed set that contains an open set. So, by the last property, it should be valid for every matrix. –  Vittorio Patriarca Jun 2 '12 at 21:20
I think you want say $R^{n^2}$ and not $R^{n}$. –  H. Kabayakawa Jun 2 '12 at 21:30