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.

Question

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.

coolng90qo · Accepted Answer

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.