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

Grayson Pierce

Grayson Pierce

Answered

2022-07-19

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

coolng90qo

coolng90qo

Expert

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.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get your answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?