Grayson Pierce

2022-07-19

Let $\mathbb{F}$ be a field and suppose $a=\left({a}_{1},\dots ,{a}_{n}\right)\in {\mathbb{F}}^{n}$ and $b=\left({b}_{1},\dots ,{b}_{n}\right)\in {\mathbb{F}}^{n}$ are such that $\sum _{k=1}^{n}{a}_{k}{b}_{k}=1$. 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 $\mathbb{F}$ 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

Expert

Yes, it is always possible. It should be straightforward that with the given information, we can always complete $\left({a}_{1},{a}_{2},\cdots ,{a}_{n}\right)$ to a full-rank matrix A with $Ab={\stackrel{^}{e}}_{1}$ (where ${\stackrel{^}{e}}_{1}$ 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 $\left\{{v}_{2},\cdots ,{v}_{n}\right\}$ that are linearly independent in the space ${b}^{\perp }=\left\{x:{x}^{T}b=0\right\}$ to serve as the remaining n-1 rows. This will always be possible as $\mathrm{dim}\left({b}^{\perp }\right)=n-1$. Since ${a}^{T}b=1\ne 0$, it will follow that $\left\{a,{v}_{2},\cdots ,{v}_{n}\right\}$ is linearly independent (why).
From here, we just need to choose the columns ${b}_{i}$ of B so that $A{b}_{i}={\stackrel{^}{e}}_{i}$. As A is invertible, this equation always has a solution.

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