We know that for a given measure (Lebesgue, or a probability measure) for the whole set of matrices $\mathbb{R}^{n\times n}$, the set of singular ones are of measure zero. However, let $S_n\subset\mathbb{R}^{n\times n}$ be the set of $n\times n$ singular matrices with real entries. If we define a nonzero measure for $S_n$ (a new probability measure for example), what can we say about the measure of the matrices with nullity (dimension of kernel) 1? Is it possible to conclude that the set of singular matrices with nullity 1 have measure 1? (among the singular ones)
I think that this is the case, since the set of matrices of nullity $n$, is just the zero matrix, so this set must be of measure 0. From this point, I don't think it should be different for matrices of nullity e.g. $n-1$: its hard for me to think that the measure distributes nonzero fractions of it between nullity $n-1$, $n-2$, etc… Moreover, if you generate singular random matrices (with normally distributed components, constrained to the $det(M)=0$ condition, for example), you get matrices with nullity 1 most of the times.
One final (bonus) question: Would the answer change, if we restrict to singular and symmetric matrices?
Best Answer
Yes.
For every $(i,j) \in \lbrace 1, ..., n \rbrace^2$, consider the application $f_{ij} : \mathcal{M}_n(\mathbb{R}) \rightarrow \mathbb{R}$ where for every matrix $M \in \mathcal{M}_n(\mathbb{R})$, $f_{ij}(M)$ is the value of the determinant of the submatrix obtained by deleting the $i-$th row and the $j-$th column of $M$.
Then the set of matrices that have a nullity of dimension $\geq 2$ is exactly $$\bigcap_{(i,j) \in \lbrace 1, ..., n \rbrace^2} f_{ij}^{-1}(0)$$
The $f_{ij}$ being polynomial, and the vanishing set of any polynomial having Lebesgue measure zero, you get that the set of matrices that have a nullity of dimension $\geq 2$ has measure zero in the set of matrices that have a nullity of dimension $1$.