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Question: Show that $\mathfrak{sl}_3(\mathbb{C})$ and $\mathfrak{s}_3(\mathbb{C})$ are ideals in $\mathfrak{gl}_3(\mathbb{C})$ and $\mathfrak{gl}_3(\mathbb{C}) = \mathfrak{sl}_3(\mathbb{C}) \oplus \mathfrak{s}_3(\mathbb{C})$.

My attempt: For all $A \in \mathfrak{gl}_3(\mathbb{C})$. If $\text{tr}(A) = 0$ then $A \in \mathfrak{sl}_3(\mathbb{C})$; else $A = \frac{\text{tr}(A)}{3} I + (A - \frac{\text{tr}(A)}{3} I)$ with $\text{tr}(A - \frac{\text{tr}(A)}{3} I) = 0$, i.e., $A - \frac{\text{tr}(A)}{3} I \in \mathfrak{sl}_3(\mathbb{C})$. Clearly $\mathfrak{sl}_3(\mathbb{C}) \cap \mathfrak{s}_3(\mathbb{C}) = \{0\}$. Hence $\mathfrak{gl}_3(\mathbb{C}) = \mathfrak{sl}_3(\mathbb{C}) \oplus \mathfrak{s}_3(\mathbb{C})$.

Is my proof true and how to prove that $\mathfrak{sl}_3(\mathbb{C})$ and $\mathfrak{s}_3(\mathbb{C})$ are ideals? Thank all!

Minh Nguyễn Hoàng
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    What is $\mathfrak{s}_3(\mathbb{C})$? – Angina Seng Jul 10 '19 at 03:23
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    Presumably $\mathfrak{s}_3(\Bbb{C})$ stands for the 1-dimensional space of 3x3 scalar matrices? To prove that they are ideals you need the rule $tr(AB)=tr(BA)$ for all square matrices $A,B$. If you have never seen that fact, prove it first! – Jyrki Lahtonen Jul 10 '19 at 03:25
  • @LordSharktheUnknown $\mathfrak{s}_3(\mathbb{C})$ is a Lie subalgebra of all matrix has form $aI$ with $a \in \mathbb{C}$ and $I$ is the unit matrix. – Minh Nguyễn Hoàng Jul 10 '19 at 03:26
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    Show $[A,B]$ has trace zero. Also show that $[A,B]=0$ when $A$ is a scalar matrix. – Angina Seng Jul 10 '19 at 03:28
  • @LordSharktheUnknown all matrix in $\mathfrak{sl}_3(\mathbb{C})$ and $\mathfrak{s}_3(\mathbb{C})$ are $3 \times 3$ square matrix, so we have $tr(AB) = tr(BA)$ and $[A,B] = AB - BA$ has trace zero. – Minh Nguyễn Hoàng Jul 10 '19 at 04:38
  • Your proof of the vector space decomposition is fine by the way. All the comments will lead you to show the summands are ideals. – Torsten Schoeneberg Jul 10 '19 at 04:39
  • @TorstenSchoeneberg We know that all matrix in $\mathfrak{sl}_3(\mathbb{C})$ and $\mathfrak{s}_3(\mathbb{C})$ are $3 \times 3$ square matrix, so $tr(AB) = tr(BA)$ and applies it to the commutator $[A,B] = AB - BA$ then we get $tr([A,B]) = 0$. Is that enough to conclude that $\mathfrak{sl}_3(\mathbb{C})$ and $\mathfrak{s}_3(\mathbb{C})$ are ideals? – Minh Nguyễn Hoàng Jul 10 '19 at 04:42
  • Would be great to use another notation, it's hard to distinguish $\mathfrak{sl}()$ and $\mathfrak{s}()$ (the latter is not very usual notation). – YCor Jul 10 '19 at 13:55

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