The Center of $\operatorname{GL}(n,k)$

Question

The given question:

Let $k$ be a field and $n \in \mathbb{N}$. Show that the centre of $\operatorname{GL}(n, k)$ is $\lbrace\lambda I\mid λ ∈ k^∗\rbrace$.

I have spent a while trying to prove this and have succeeded if $ k \subseteq \mathbb{R}$. So I imagine there is a nicer way to go about this. I have seen people saying take matrices where every element except one is zero in proving similar results but such matrices are not in $\operatorname{GL}(n,k)$ as they are not invertible.

What I have done.

I have said let $B \in$ Center then $B$ commutes with everything in $\operatorname{GL}(n,k)$. So take a permutation matrix, $P$ which swaps rows $i,j$ $P \neq I$. From this you can deduce that $B^T = B$ and that $B_{ii} = B_{jj}$. We can use the fact that $B$ is symmetric to find an orthogonal diagonalisation of $B$ by the spectral theorem.

This gives $QBQ^T = D \rightarrow B=D$. And as $B_{ii} = B_{jj}$ we can say $B = k * I$.

But the spectral theorem requires $B$ to be a real matrix.

How do I prove this for a general $k$?.

Answer 1

Let $A$ belong to the center of $GL(n,k)$.

Fix $1\leq i\neq j\leq n$.

Denote $E_{i,j}$ the matrix with $1$ as $(i,j)$ coefficient, and $0$ elsewhere.

Then $I_n+E_{i,j}$ belongs to $GL(n,k)$ (as requested below, a proof of this is that, like every transvection matrix, $I_n+E_{i,j}$ has determinant $1$).

So $$ (I_n+E_{i,j})A=A(I_n+E_{i,j})\quad \Leftrightarrow\quad E_{i,j}A=AE_{i,j}. $$ Now compute the appropriate coefficients in the latter.

Answer 2

Let $A$ be a matrix in the center of $\operatorname{GL}(n,k)$. Then it commutes with all invertible matrices. So look at invertible matrices $B$ with an easy structure (for example, having only entries 0 or 1, and as few 1's as possible) and translate the equation $AB = BA$ into conditions on the entries of $A$.