Let $(A\_{1},A\_{2}, \ldots,A\_{k})$ be $k$ matrices in $M\_{n}(\mathbb{R})$.


Is there an algebraic formula, as a generalization of "Determinant" for $k=1$, to compute the joint spectrum of $(A\_{1},A\_{2}, \ldots,A\_{k})$?


The joint spectrum is the union of righ and left joint spectrum. The later is defined as follows:


**Right joint spectrum=** All $(\lambda\_{1}, \lambda\_{2}, \ldots, \lambda\_{k})$ such that the right ideal generated by $(A\_{i} - \lambda\_{i}I)\_{i=1}^{k}$ is not the whole $M\_{n}(\mathbb{R}).$


Moreover, motivated by Perron–Frobenius theorem we ask:



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> Assume that all enries of all matrices $(A\_{1},A\_{2}, \ldots,A\_{k})$ are positive. Is there an element $(\lambda\_{1}, \lambda\_{2}, \ldots, \lambda\_{k})$ of joint spectrum with $\lambda \geq 0$?
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> 
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How to compute the joint spectrum?
Let us write $T=M\_n(\mathbb{R})$. We have a polynomial map $f:\mathbb{R}^k\times T^k\to Hom\_{\mathbb{R}}(T^k,T)$ which sends $((\lambda\_i),(A\_i))$ to the linear map $$(B\_i)\mapsto \sum\_i B\_i(A\_i-\lambda\_i).$$
It holds that $f((\lambda\_i),(A\_i))$ is not surjective if and only if $(\lambda\_i)$ is in the left joint spectrum of $(A\_i)$.
The lack of surjectivity of $f((\lambda\_i),(A\_i))$ can be written in terms of the vanishing of some minors. This describes the left joint spectrum as a Zariski closed subset of an affine space, where the coefficients of the polynomials are given by polynomials in the entries of $A\_i$. A similar thing can be done of course to the right joint spectrum, and since the union of Zariski closed subsets is Zariski closed, we get the result we want.


