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 F and for all ( i , j ) E n 2 ( E n = { 1 , 2 , . . . , n }), P i j : V V is a linear map. Suppose moreover that...
1) ( i 1 , j 1 , i 2 , j 2 ) E n 4
2) { v j V } 1 j n obey the following set of equations
i E n : j = 1 n P i j v j = 0.
Then j E n
( det P ) v j = 0
where P is the n × n matrix with entries P i j and det P 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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Answers (1)

Abigail Palmer
Answered 2022-06-27 Author has 30 answers
This can be derived from the Cayley-Hamilton theorem for commutative rings.Condition 1 says that the unital F-algebra R generated by the P i j is a commutative ring.The Cayley-Hamilton theorem says that for any P R n × n ,, if we define a polynomial p over R by p ( λ ) = det ( λ P ) ,, then p ( P ) = 0 R n × n ..Writing for the action of R n × n on V n ,, if P v = 0 we get
0 v = p ( P ) v = p ( 0 ) v = det ( P ) v .
where in the second equality I am using the fact that the P v , P 2 v , terms vanish.

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