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I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:End(\Bbb R^{2n})\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ is a $C^\infty$ map. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f_n(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f_n)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f_n)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_n\operatorname{img}(g\circ f_n)=\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:A\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ where $A\subseteq End(\Bbb R^{2n})$ are the endomorphisms of rank $2$ is $C^\infty$. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f_n(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f_n)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f_n)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_n\operatorname{img}(g\circ f_n)=\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:End(\Bbb R^{2n})\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ is a $C^\infty$ map. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:End(\Bbb R^{2n})\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ is a $C^\infty$ map. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f_n(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f_n)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f_n)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_n\operatorname{img}(g\circ f_n)=\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:End(\Bbb R^{2n})\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ is a $C^\infty$ map. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f(x,y)= \operatorname ({ker}\circ P(x),y)$. The rank of $D(g\circ f)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.

I assume $P$ varies analytically as in the body of the question, I do not know how to tackle the case where $P$ is just differentiable, I also read rank 2 as real rank though of course the problem is even easier if the rank is complex.

The map $\operatorname {ker}:End(\Bbb R^{2n})\rightarrow\operatorname {Gr}_{2n-2}\Bbb R^{2n}$ is a $C^\infty$ map. Clearly the rank of $D(\operatorname {ker}\circ P)$ is less than or equal to 1. Consider the tautological bundle over the Grassmanian $T$ and the projection map $g:T\rightarrow \Bbb R^{2n}$. Since $T$ is locally trivial and second countable, we have $T$ is a countable union of spaces diffeomorphic to $\operatorname {Gr}_{n-2}\Bbb R^{2n}\times \Bbb R^{2n-2}=M_n$. Consider $f:\Bbb R\times\Bbb R^{2n-2} \rightarrow M_n$, $f(x,y)= (\operatorname {ker}\circ P(x),y)$. The rank of $D(g\circ f)$ is less than or equal to $2n-1$, so by Sard's theorem we have $\operatorname{img}(g\circ f)$ is measure zero in $\Bbb R^{2n}$. So $\bigcup_{t \in [0,1]} \mathrm{ker}\, P(t)$ is a countable union of measure zero sets hence its complement is non-empty.