Let F be a field and suppose a = ( a 1 , … ,...

Grayson Pierce

Grayson Pierce



Let F be a field and suppose a = ( a 1 , , a n ) F n and b = ( b 1 , , b n ) F n are such that k = 1 n a k b k = 1. Is it always possible to find n × n matrices A and B such that the first row of A is a, the first column of B is b, and AB=I?
To me the answer is clearly yes but I am having some trouble proving this. I am fairly certain it is not needed but F is algebraically closed.
For those wondering what the context is for this problem, I am trying to determine whether two subalgebras are conjugate and I can construct an inner automorphism provided this fact always holds.

Answer & Explanation




2022-07-20Added 14 answers

Yes, it is always possible. It should be straightforward that with the given information, we can always complete ( a 1 , a 2 , , a n ) to a full-rank matrix A with A b = e ^ 1 (where e ^ 1 is the first column of the identity, and is often referred to as the first standard basis vector). Indeed, we just need to find n-1 vectors { v 2 , , v n } that are linearly independent in the space b = { x : x T b = 0 } to serve as the remaining n-1 rows. This will always be possible as dim ( b ) = n 1. Since a T b = 1 0, it will follow that { a , v 2 , , v n } is linearly independent (why).
From here, we just need to choose the columns b i of B so that A b i = e ^ i . As A is invertible, this equation always has a solution.

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