Suppose that $P(t)$ is a one-parameter family of rank 2 self-adjoint projections on $\mathbb{C}^n$ that vary analytically in the real parameter $t \in [0,1]$. I claim that there must exist a vector $x \in \mathbb{C}^n$ such that $P(t)x \neq 0$ for all $t$. 


In other words, I am hoping to prove that $\bigcup\_{t \in [0,1]} \mathrm{ker}\, P(t) \neq \mathbb{C}^n$. This seems like it is essentially a space-filling curve type argument, and hence the requirement that $P(t)$ be differentiable is probably important (in my example, $P(t)$ is analytic in $t$). Does anyone know a reference that would provide a simple proof of this claim? 



A differentiable one-parameter family of codimension 2 subspaces of $\mathbb{C}^n$ cannot fill $\mathbb{C}^n$, right?
Let $Q(t):\mathbb C^n\to\mathbb C^n$ be the orthogonal projection onto $\text{ker}(P(t))$.
Then $t\mapsto Q(t)$ is as differentiable as $P$ was. Now the mapping $(t,x)\mapsto Q(t)(x)$ has rank at most $2n-2+1<2n$. Apply Sard's theorem: The set of regular values is Lebesgue nearly everything. Take a regular value $y$. It cannot be in the image, and thus is in no kernel of $P(t)$ for any $t$.


