 # Let V be a vector space over a field <mi mathvariant="double-struck">F and for all ( Yahir Crane 2022-06-26 Answered
Let $V$ be a vector space over a field $\mathbb{F}$ and for all $\left(i,j\right)\in {\mathbb{E}}_{n}^{2}$ (${\mathbb{E}}_{n}=\left\{1,2,...,n\right\}$), ${P}_{ij}:V\to V$ is a linear map. Suppose moreover that...
1) $\mathrm{\forall }\left({i}_{1},{j}_{1},{i}_{2},{j}_{2}\right)\in {\mathbb{E}}_{n}^{4}$
2) $\left\{{v}_{j}\in V{\right\}}_{1\le j\le n}$ obey the following set of equations
$\mathrm{\forall }i\in {\mathbb{E}}_{n}:\phantom{\rule{thinmathspace}{0ex}}\sum _{j=1}^{n}{P}_{ij}{v}_{j}=0.$
Then $\mathrm{\forall }j\in {\mathbb{E}}_{n}$
$\left(detP\right){v}_{j}=0$
where $P$ is the $n×n$ matrix with entries ${P}_{ij}$ and $detP$ has the same combinatorial structure as the usual determinant, yet with all multiplications replaced by map compositions.
Question: What name(s) is given to this type of result in the mathematical literature? Are there short, elegant proofs?
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This can be derived from the Cayley-Hamilton theorem for commutative rings.Condition 1 says that the unital $\mathbb{F}$-algebra $R$ generated by the ${P}_{ij}$ is a commutative ring.The Cayley-Hamilton theorem says that for any $P\in {R}^{n×n},$, if we define a polynomial $p$ over $R$ by $p\left(\lambda \right)=det\left(\lambda -P\right),$, then $p\left(P\right)=0\in {R}^{n×n}.$.Writing $\ast$ for the action of ${R}^{n×n}$ on ${V}^{n},$, if $P\ast v=0$ we get
$0\ast v=p\left(P\right)\ast v=p\left(0\right)\ast v=det\left(-P\right)\ast v.$
where in the second equality I am using the fact that the $P\ast v,{P}^{2}\ast v,\dots$ terms vanish.

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