For square matrices whose product is defined, $\operatorname{tr}(A_1A_2\cdots A_n)$ is unchanged by a cyclic permutation of the factors. In particular, every <commutator> has trace zero.

Cyclic property of the trace

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{title2=$[x,y]$}
{wiki}

The Lie bracket is the product in a <Lie algebra>. For matrix Lie algebras it is the <commutator> $[X,Y]=XY-YX$.

= Lie brackets
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Lie bracket

With
$$
J=\begin{pmatrix}0&1\\-1&0\end{pmatrix},
\qquad
A=\begin{pmatrix}p&q\\r&s\end{pmatrix},
$$
the equation $AJ+JA^T=0$ reduces to $p+s=0$. Hence
$$
\mathfrak{sp}_2
=\left\{\begin{pmatrix}p&q\\r&-p\end{pmatrix}:p,q,r\in\mathbb C\right\}
=\mathfrak{sl}_2
$$
as matrix <Lie algebras>, with the same <commutator> bracket.

Solved by gpt-5.6-sol high.

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 102 / 3 / iii / Solution

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{title2=$\mathfrak{sl}_n$}
{wiki=Special_linear_Lie_algebra}

The special linear Lie algebra $\mathfrak{sl}_n$ consists of the trace-zero $n\times n$ matrices with the <commutator> bracket.

Special linear Lie algebra

{title2=$[A,B]$}
{wiki=Commutator}

The commutator of two elements of an associative algebra is $[A,B]=AB-BA$. Matrix Lie algebras use the commutator as their <Lie bracket>.

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 Incoming links: Commutator