I believe that it is but how can one show this? This is for study so any help would be great and appreciated.
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What do you get if you multiply an element of SL(2, C) with, say, a matrix of determinant 2? – Justin Barhite Nov 22 '19 at 11:54
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@JustinBarhite I think you are confusing the linear groups $SL(n)$ and $GL(n)$ endowed with usual product, and the Lie algebras $\mathfrak{sl}(n)$ ($n\times n$ matrices with trace $0$) and $\mathfrak{gl}(n)$ ($n\times n$ matrices) endowed with bracket $[A,B]=AB-BA$ – Nov 22 '19 at 12:03
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Possible duplicate of Derived Algebra of Sl(n,C) – Dietrich Burde Nov 22 '19 at 12:04
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So we have $[\mathfrak{gl}_n(K),\mathfrak{gl}_n(K)]=\mathfrak{sl}_n(K)$, which is the derived ideal, hence an ideal, see the duplicate. – Dietrich Burde Nov 22 '19 at 12:05
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Yes I have no idea how I did not think of that..... thank you though @DietrichBurde – NumberFrog Nov 22 '19 at 12:17
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In every Lie algebra $L$ is the commutator subalgebra $[L,L]$ an ideal. Since it is well-known that $$ [\mathfrak{gl}_n(K),\mathfrak{gl}_n(K)]=\mathfrak{sl}_n(K), $$ the claim follows.
Dietrich Burde
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Yes: $[A,B]\in\mathfrak{sl}(n, \Bbb F)$ not only when $B\in\mathfrak{sl}(n, \Bbb F)$ and $A\in\mathfrak{gl}(n, \Bbb F)$ (which would be your goal), but also for all $A,B\in\mathfrak{gl}(n, \Bbb F)$. This is due to the identity $\operatorname{tr}(AB)=\operatorname{tr}(BA)$.