# Finding Common Eigenvalues and Eigenvectors via Commutative Algebra?

I am interested in the following question:

Question: Let $$H(\lambda)$$ be a Hermitian matrix whose entries are polynomials of a parameter $$\lambda\in\mathbb C$$. Consider all pairs of $$(\lambda, \lambda')\in \mathbb C^2$$ that satisfies a polynomial relation $$F(\lambda, \lambda')=0$$. For what $$(\lambda, \lambda')$$ among these pairs, the matrices $$H(\lambda)$$ and $$H(\lambda')$$ has a common eigenvalue and a corresponding common eigenvectors (up to constant multiples)?

I want to know an algorithm to answer this question.

Motivation: I want to solve a certain type of PDE on a region $$\Omega$$, which is a disjoint union of $$\Omega_1, \Omega_2$$. The most general solution of PDE in each subregion is known. To construct the solution on the whole $$\Omega$$, I have to impose the continuity on the boundary of $$\Omega_1$$ and $$\Omega_2$$. This matter reduces to the above problem!

The most brute force approach to this problem would be listing all eigenvalues and eigenvectors, and finding the common one. However, since the matrix size is large, the characteristic polynomial is of large degree. Then, one cannot explicitly find the roots of the polynomial!

I suspect that there is a more algebraic approach (using commutative algebra, Grobner basis, etc) to this problem. Although I have bare knowledge of commutative algebra, I know that the existence of common roots of two polynomials can be easily determined by computing the resultants. Hence, if one only cares about the eigenvalues, then the resultant would be an answer. But what about eigenvectors also?

Remark: I have a Mathematica code for the explicit form of $$H(\lambda)$$ and $$F(\lambda, \lambda')$$.

• What is the relationship between $F(\lambda,\lambda')$ and $H(\lambda)$? Mar 25, 2021 at 14:54
• @BenGrossmann It is assumed to be independent. Does it help if there is a particular relationship between $F(\lambda, \lambda')$ and $H(\lambda)$? The answer assuming this is also welcome! Mar 25, 2021 at 15:22

One idea is as follows: let $$n$$ denote the size of $$H(\lambda),H(\lambda')$$. Introduce a variable $$x$$, which will stand for the common eigenvalue of $$H(\lambda)$$ and $$H(\lambda')$$. The matrices $$H(\lambda), H(\lambda')$$ have common eigenvalue $$x$$ with some associated common eigenvector $$v$$ if and only if
$$\operatorname{rank} \pmatrix{H(\lambda) - xI\\ H(\lambda') - xI} \leq n-1.$$
Note that a matrix $$M$$ (with at least $$n$$ rows and columns) will have rank at most $$n-1$$ if and only if all of its submatrices of size $$n \times n$$ have determinant zero. Thus, the condition above can be expressed by setting all $$\binom{2n}{n}$$ such determinants equal to zero.
If we introduce the additional condition $$F(\lambda,\lambda') = 0$$, then we can characterize the desired set of values $$(x,\lambda,\lambda')$$ as the solution to a system of polynomial equations. From there, one could apply some standard approach.
• @eigenvalue It's a useful fact. If one is comfortable introducing complex conjugates, then it can also be useful to note that $M$ has full column-rank if and only if $\det(M^*M) \neq 0$ (where $M^*$ denotes the conjugate-transpose of $M$). Mar 25, 2021 at 15:55
• @eigenvalue Sometimes using the complex-conjugate leads to trouble. In particular, using the complex conjugate on a complex polynomial doesn't necessarily leave you with a complex polynomial. For instance, the fundamental theorem of algebra guarantees that every polynomial as a root, but the function $$f(\lambda) = \lambda \cdot \bar \lambda + 1$$ has no root. Mar 25, 2021 at 23:46