Continuity of eigenvectors

Question

Let $\mathbb{C} \ni z \mapsto M(z)$ be a square matrix depending holomorphically on a parameter $z$ with the property that $\operatorname{dim}\ker(M(z)))=1$ for $z $ away from a discrete set $D \subset \mathbb{C}$ and $\operatorname{dim}\ker(M(z)))\ge 1$ for $z \in D.$

I ask: Is it always possible to choose a continuous vector $\mathbb{C} \ni z \mapsto v(z)$ such that $M(z)v(z)=0?$

Answer 1

Yes. Let the size of your matrix be $n$. Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$. Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that $M(z)u(z)=0$ by solving the system of $n-1$ linear equations in $n-1$ variables. This requires only arithmetic operations on the entries of $M$, so $u(z)$ will be meromorphic on $\mathbb{C}$. Let $D$ be the divisor of poles of $u$. According to a theorem of Weierstrass there is an entire function $f$ having zeros at $D$. Then $v(z)=f(z)u(z)$ is a solution to your problem, which is not only continuous but holomorphic.