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

  • 2
    $\begingroup$ 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 $z$. $\endgroup$ 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.