Continuity of eigenvectors

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 $$M(z)v(z)=0?$$

• The second requirement dim ker$(M(z))\ge 1$ for $z\in D$ follows from the first: Otherwise $M(z)$ would be invertible in $z$ and hence in a neighbourhood of $z$. Jul 23 '20 at 9:49

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.