Intersection of kernels of two matrices with nonlinear coefficients Consider $A$ a $n\times n$ skew-symmetric matrix and $B$ a $n\times n$ symmetric matrix, the coefficients of which are (independent) nonlinear equations in $t_1,\dots,t_n$. For example, the element $(1,1)$ of $A$ could be $a_{11}=\sin(3t_1)+\sqrt{2}\sin(t_2)$.
Is it true to say that, in general, there is no $(t_1,\dots,t_n)\in\mathbb{R}^n$ and $0\neq x\in\mathbb{R}^n$ such that $Ax=0$ and $Bx=0$?
I would think that this is false because necessarily, $A,B$ are singular, so both matrix equations give at most $2(n-1)$ scalar independent equations for $2n$ unknowns $(t_1,\dots,t_n)$ and $(x_1,\dots,x_n)$, corresponding to an underdetermined system. But I'm not sure.
 A: We work over $\mathbb{C}$ ( not over $\mathbb{R}$) and we assume that $A=[a_{i,j}],B=[b_{i,j}]$ where $a_{i,j},b_{i,j}$ are "generic" (to get an idea: random chosen polynomials) algebraic functions of $(t_1,\cdots,t_n)$. 
You want that there is $x\not= 0$ s.t. $Ax=Bx=0$. Then necessarily $\det(B)=0$. Then $V=\{(t_i)|\det(B)=0\}$ is an algebraic set of dimension $n-1$. For a generic $(t_i)\in V$, there is a sole solution of $Bx=0$ in the form $x((t_i))=[1,a_2((t_i)),\cdots,a_n((t_i))]^T$ where $(t_i)\in V$ and the functions $a_2,\cdots,a_n$ are unique. It remains to consider the equality $Ax((t_i))=0$, that is $n$ equations in $(t_i)$.
Case 1. $n$ is odd. Then, for any $(t_i)\in V$, $\det(A)=0$ and there are only $n-1$ algebraically independent relations linking the elements of $V$. Generically, there are a finite number of solutions.
EDIT. My reasoning when $n$ is even was false and the anderstood's comment is correct. Moreover, I edited also the example when $n=3$.
According to Fitzgerald, too much of anything is bad, but too much Champagne is just right -except when doing maths-.
Case 2. $n$ is even. Then $x((t_i))\not=0$ and $x((t_i))^TAx((t_i))=0$, that is a non-trivial combination of our $n$ equations in $((t_i))$. Generically $n-1$ equations are enough (in particular, they imply that $\det(A)=0$). As when $n$ is odd, generically, there are a finite number of solutions (a countable number if the functions $(a_{i,j}),(b_{i,j})$ are holomorphic and not algebraic).
Example with $n=3$. Let $A=\begin{pmatrix}0&-w&v\\w&0&-u\\-v&u&0\end{pmatrix},B=\begin{pmatrix}a&b&c\\b&d&e\\c&e&f\end{pmatrix}$ and let $s=[df-e^2,-bf+ce,be-cd]^T,p=[a,b,c]^T,q=[u,v,w]^T$. Under the condition, $\det(B)=<p,s>=0$ the solution (unique up to a factor if $rank(B)=2$) is $s\not=0$, satisfying also $As=s\wedge q=0$. 


*

*$q\not=0$. Then $s$ and $q$ are parallel vectors ($2$ relations) whence a total of three relations linking $t_1,t_2,t_3$; finally we obtain generically a finite number of solutions.

*$q=0$ ($3$ relations) whence a total of four relations; generically there are no solutions.
