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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.

Thanks in advance.

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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.

You can prove that it not only doesn't have a center, but it also doesn't have any non-trivial ideals.

The smallest subalgebra of $gl_n$ having scalar matrices as its center will be the (abelian) subalgebra, consisting of scalar matrices.

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