For fixed matrices $A$ and $B,$ can we have $\ker(A+tB)\neq\ker(A+sB)$ for some reals $s$ and $t?$ Let $A,B$ be matrices such that $\ker(A+tB)$ is a 1 dimensional space for any $t$ in the interval $[-\epsilon,\epsilon]$.
Does there exist such a pair of matrices $A,B$ that satisfy the above in addition to having $\ker(A+tB)\neq\ker(A+sB)$ for some reals $s,t\in(-\epsilon,\epsilon)$?
Let $v_t$ be a unit vector in $\ker(A+tB)$. I would like to find
$\lim_{t\rightarrow0}\frac{v_t-v_0}{t}$.
Because the dimension of the kernel is 1, there are only 2 such unit vectors, however one still needs to be careful to pick the v's so they are close together so that the limit converges. Assume this is done when taking the above limit.
So my two questions can be summarized as follows: Can the kernel move and if so, what is the rate of change of a unit vector in the kernel?
 A: A simple example where the kernels are different.
$$A=\begin{pmatrix}1&0\\1&0\end{pmatrix},B=\begin{pmatrix}0&1\\0&1\end{pmatrix}$$ then $$\ker(A+Bt)=\left\langle\begin{pmatrix}t\\-1\end{pmatrix}\right\rangle$$

The rest only covers how to define $v_t$ to be continuous on $t.$
Since $\det (A+tB)=0$ is a polynomial which is zero for infinitely many $t,$ then it must be the zero polynomial.
This means, as a matrix over the field, $K=\mathbb R(t),$ there must be a non-zero vector $\mathbf v(t) \in K^n$ with $(A+Bt)\mathbf v(t)=0.$ We can make the entries of $\mathbf v(t)$ be polynomials (by multiplying by all the denominators,) and then remove common polynomial factors.
So $\mathbf v(t)$ becomes, as a function on $t\in \mathbb R,$ a continuous function which never takes the value $\mathbf 0.$ So you can use: $$v_t=\frac{\mathbf v(t)}{\|\mathbf v(t)\|}.$$

The same argument applies if the matrices are complex, just replacing $\mathbb R$ With $\mathbb C.$

Other incomplete ideas.
The characteristic polynomial of $A+Bt\in M_n(K),$ $$p(\lambda)=\det(\lambda I-(A+Bt))$$ is of the form:
$$\lambda^n +\sum_{k=1}^{n-1}q_k(t)\lambda^k$$ where $\deg q_k\leq n-1.$ We can skip $k=0$ because $\det(A+Bt)=0.$
Thus the minimal polynomial $m(\lambda)=\lambda m_0(\lambda)$ is also of this form, I believe.
Then $m_0(A+Bt)$ has a column space of dimension $1.$ Otherwise, we’d have two linearly independent $\mathbf v_1(t),\mathbf v_2(t)$ and thus two vectors in the kernel, for at least some real $t,$ it would seem.
Can $\mathbf v_i(t)$ be linearly dependent for all $t$ but independent in $K^n?$

We can limit to the cases when $A$ is in Jordan normal form.
In the case of $2\times 2$ matrices, this means you can restrict to $$A_1=\begin{pmatrix}0&0\\0&1\end{pmatrix},\\A_2=\begin{pmatrix}0&1\\0&0\end{pmatrix}$$
Now, let $B=(b_{ij})_{i,j=1}^2.$
The characteristic polynomial for $A_1+Bt$ is $$\lambda^2-(t(b_{11}+b_{22})+1)\lambda +t^2\det B+b_{11}t.$$
So you need $b_{11}=\det B=0.$ This means $B$ must either have a zero first row or zero first column.
If $B$ has a zero first column, then the kernel,s are all the same, generated by $(1,0).$
But in the other case, you need:
$$t(b_{21}x+b_{22}y)+y=0,$$
If $y=t,$ you get:
$$x=\frac{-1-b_{22}t}{b_{21}}.$$

The case for $A_2+Bt$ requires $b_{21}=0$ and $b_{11}b_{22}=0.$
Again, the kernels are equal when $b_{11}=0.$
You get a kernel generated by $y=t, x=-\frac{1+b_{12}t}{b_{11}}.$
So for $2\times 2$ matrices, $\mathbf v(t)$ can be made linear.

You might be able to generate similar examples in $3\times 3$ matrices. I wouldn’t be totally surprised if it was always linear, but it seems unlikely on first blush. It is certainly the case that we can find $\mathbf v(t)$ where each component is of degree at most $n-1.$
