# What is the smallest Lie subalgebra of ${{\frak{gl}}_{n}}(\mathbb{R})$ whose center is the set of $(n \times n)$-scalar matrices?

We know that the center of the Lie algebra ${{\frak{gl}}_{n}}(\mathbb{R})$ of all $(n \times n)$-matrices is the Lie subalgebra of all $(n \times n)$-scalar matrices. The Lie algebra ${{\frak{sl}}_{n}}(\mathbb{R})$ of all $(n \times n)$-matrices with zero trace has the same center.

My question is:

What is the ‘smallest’ Lie subalgebra of ${{\frak{gl}}_{n}}(\mathbb{R})$ whose center is the set of all $(n \times n)$-scalar matrices?

I know that ${{\frak{gl}}_{n}}(\mathbb{R})$ has non-comparable Lie subalgebras, so I cannot define the term ‘smallest’ clearly. Still, I hope that the context itself is clear.

I think $sl_n$ does not have a center. If a scalar matrix is in $sl_n$, then its trace must be $0$, so it can only be a zero matrix.
The smallest subalgebra of $gl_n$ having scalar matrices as its center will be the (abelian) subalgebra, consisting of scalar matrices.