A question on Grassmannian Let $V$ be the space of $4$ by $4$ Hermitian matrices, that
is a vector space of dimension $16$ over $\mathbb{R}$. Is the uniform
measure of 
$$
\left\{ W\in Gr\left(5,V\right):W \text{contains no nonzero Hermitian matrix with at least 2 eigenvalues that are 0}\right\} 
$$
equal to $0$?
Does any of the experts know or have the answer to this question?
 A: It appears that the answer is 'no', based on the answer to the OP's previous question, A question on eigenvalues.  In the answers to that question, it is pointed out that there is a 1965 paper by Adams, Lax, and Phillips that implies that there exists a 5-dimensional subspace $W_0\subset V$ such that its nonzero elements are, in fact, nonsingular (which is considerably stronger than what the OP required).
If you let $\hat C\subset V$ denote the $12$-dimensional cone of Hermitian matrices with at least two zero eigenvalues, then the projectivization of $\hat C$ is a closed algebraic subvariety $C\subset\mathbb{P}(V)\simeq\mathbb{RP}^{15}$ of dimension $11$.  By construction $\mathbb{P}(W_0)\subset \mathbb{P}(V)$ does not meet $C$.  The set of subspaces $W\in\mathrm{Gr}(5,V)$ such that $\mathbb{P}(W)\cap C=\emptyset$ is therefore nonempty and it is clearly open in $\mathrm{Gr}(5,V)$ (because $C$ is closed).  Therefore, in particular, it has nonzero uniform measure.
NB:  The OP asked what I meant by 'near' in my comment above.  If one fixes an inner product $q$ on $V$, say the obvious $\mathrm{U}(4)$ invariant one (but any positive definite inner product will do), then there is induced on each $\mathrm{Gr}(k,V)$ a natural metric, unique up to scale, that is invariant under the the orthogonal group of $q$.  By 'near', I meant 'close in such a metric'.
